Are there incongruent pythagorean triangles with the same perimeter and same area?












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I found there are two incongruent isosceles triangles with integer sides and areas, where both have same perimeter, same area.



I looked around Dickson's History of Number Theory but couldn't find where the right triangle version is treated. [I thought if a nonexistence proof was simple it would pop up in my search, but found none.]



It may be simple to show none exist, but I had no luck, only filled few notebook pages with formulas going nowhere. Reference/example/proof appreciated. Thanks.










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    7












    $begingroup$


    I found there are two incongruent isosceles triangles with integer sides and areas, where both have same perimeter, same area.



    I looked around Dickson's History of Number Theory but couldn't find where the right triangle version is treated. [I thought if a nonexistence proof was simple it would pop up in my search, but found none.]



    It may be simple to show none exist, but I had no luck, only filled few notebook pages with formulas going nowhere. Reference/example/proof appreciated. Thanks.










    share|cite|improve this question











    $endgroup$















      7












      7








      7





      $begingroup$


      I found there are two incongruent isosceles triangles with integer sides and areas, where both have same perimeter, same area.



      I looked around Dickson's History of Number Theory but couldn't find where the right triangle version is treated. [I thought if a nonexistence proof was simple it would pop up in my search, but found none.]



      It may be simple to show none exist, but I had no luck, only filled few notebook pages with formulas going nowhere. Reference/example/proof appreciated. Thanks.










      share|cite|improve this question











      $endgroup$




      I found there are two incongruent isosceles triangles with integer sides and areas, where both have same perimeter, same area.



      I looked around Dickson's History of Number Theory but couldn't find where the right triangle version is treated. [I thought if a nonexistence proof was simple it would pop up in my search, but found none.]



      It may be simple to show none exist, but I had no luck, only filled few notebook pages with formulas going nowhere. Reference/example/proof appreciated. Thanks.







      geometry






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      edited 14 hours ago









      Gregory Nisbet

      835712




      835712










      asked 19 hours ago









      coffeemathcoffeemath

      2,9541415




      2,9541415






















          5 Answers
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          7












          $begingroup$

          Consider two right triangles, with hypotenuses $p$ and $q$ and respective acute angles $theta$ and $phi$. To see that having equal perimeter and area makes them congruent, it suffices to show that
          $$theta = phi qquadtext{or}qquad theta+phi=frac{pi}{2} tag{0}$$
          (Either makes the triangles similar, which in turn makes them congruent.)



          Equating perimeters and areas gives a system we can write as



          $$begin{align}
          p(1+sintheta+costheta) &= q(1+sinphi+cosphi) \
          p^2 sintheta costheta &= q^2 sinphi cosphi
          end{align} tag{1}$$



          Defining $u:=tan(theta/2)$, we "know" that
          $$sintheta = frac{2u}{1+u^2} qquad costheta=frac{1-u^2}{1+u^2} quadtoquad 1+costheta+sintheta= frac{2 (1 + u)}{1 + u^2}$$
          and likewise for $v:=tan(phi/2)$. Thus, $(1)$ can be rewritten as
          $$begin{align}
          pfrac{(1+u)}{1+u^2} &= qfrac{(1+v)}{1+v^2} \[4pt]
          p^2frac{u(1+u)(1-u)}{(1+u^2)^2} &= q^2frac{v(1+v)(1-v)}{(1+v^2)^2}
          end{align}tag{2}$$

          Dividing the second equation by the square of the first ...



          $$frac{u(1-u)}{1+u} = frac{v(1-v)}{1+v} quadtoquad (u-v)(uv+u+v-1)=0 quadtoquad u=v, text{ or } frac{u+v}{1-uv}=1 tag{3}$$



          Therefore, we have one of the following situations (bearing in mind that $theta/2$ and $phi/2$ are each at most $pi/4$, so that we may draw appropriate conclusions from these tangent inequalities):
          $$begin{align}
          tanfrac{theta}{2}=tanfrac{phi}{2} &quadtoquad theta=phi \[4pt]
          frac{tan(theta/2)+tan(phi/2)}{1-tan(theta/2)tan(phi/2)} = 1 &quadtoquad
          tanleft(frac{theta}{2}+frac{phi}{2}right)=tanfrac{pi}{4} quadtoquad theta+phi=frac{pi}{2}end{align} tag{4}$$

          which match the sufficient conditions in $(0)$. $square$






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            Let $S$ be the circumference and $A$ twice the area of a triangle.



            Then,
            $$a_i+b_i+sqrt{a_i^2+b_i^2}=S
            text{ and } a_ib_i=A. tag{1}$$

            After squaring, $S^2+2A-2S(a_i+b_i)=0$ and from here $a_i+b_i=S/2+A/S$. Thus, $c_i=S-(a_i+b_i)=S/2-A/S=c$, that is, triangles have the same hypotenuse.



            Then, $a_i+b_i=S-c=T$ and $a_ib_i=A$, which results in a solutions for $a_i$ and $b_i$ expressed in terms of constants $A$ and $T$. Although one of the resulting equations is quadratic, there is a symmetric pair of solutions (thank you for comments below). Hence, all sides must be the same.






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            • 3




              $begingroup$
              Really nice proof. I have a niggle with your last statement though. In general, $2$ equations with $2$ unknowns may well have multiple solutions if they are not linear. For example, the OP mentions isosceles triangles, where the triangles $(8,8,12)$ and $(6,11,11)$ have the same perimeter and area even though isosceles triangles are also parametrised by two variables. Of course, in your case you know the sum and product of $a$ and $b$, which is a particularly nice pair of equations well known to have a symmetric pair of solutions.
              $endgroup$
              – Jaap Scherphuis
              17 hours ago












            • $begingroup$
              @Jaap Scherphuis -- Thank you for your comment and clarification.
              $endgroup$
              – dnqxt
              17 hours ago












            • $begingroup$
              @Dawood ibn Kareem -- Thanks for the comment. I added in the answer the clarification made by Jaap Scherphuis.
              $endgroup$
              – dnqxt
              10 hours ago



















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            Consider a right-angled triangle with sides $a$ and $b$.

            The hypotenuse has length $c=sqrt{a^2+b^2}$.
            Its area is $ab/2$, and perimeter is $a+b+c$.



            I will allow $a,b,c$ to be any positive real numbers, not restrict them to positive integers.



            Suppose we scale the triangles such that $ab=1$, (i.e. an area of $1/2$). Is it possible to have two of these triangles that are distinct but with the same perimeter?



            We may assume that $a$ is the longer side, i.e. $a>b$, so we must have $a>1$.



            The perimeter is
            $$P(a) = a+b+c\ = a+b+sqrt{a^2+b^2}\ = a+frac{1}{a}+sqrt{a^2+frac{1}{a^2}}$$



            This is an increasing function on the interval $[1,infty)$ because its derivative w.r.t. $a$ is positive for $a>1$. This is tedious to check by hand, so I used Wolfram alpha. You can however understand why this is the case by noticing that if you increase $a$, then the rate at which $a$ increases is larger than the rate at which $1/a$ decreases, and the same holds for $a^2$ versus $1/a^2$.



            This means that there are no two values of $a$, both with $a>1$ for which you get the same perimeter.



            Bringing it back to the original problem, it means that there are no two right-angled triangles with the same perimeter and area, unless they have the same sides. Basically, given an area and a perimeter, their two equations uniquely determine the two triangle sides because the lines those equations represent are not curved enough to intersect multiple times.





            P.S. By the way, the OP mentioned that there are pairs of isosceles triangles with matching areas and perimeters, even when all sides and the area are integers. Two examples are:
            $(29,29,40)$ and $(37,37,24)$,

            and also
            $(218,218,240)$ and $(233,233,210)$.






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              The area and the perimeter uniquely define the radius of the inscribed circle because $A=frac12Pr$, and the hypotenuse because $r=frac{P}{2}-c$. That fixes both $a+b=P-c$ and $ab=2A$ so $a$ and $b$ are also unique up to permutation, QED.






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                Long Comment:



                You can at least give the simple formulae for the perimeter $P$ and area $A$ of a right angle triangle.



                If $z^2=x^2+y^2$ is a primitive Pythagorean triangle with $x$ being the base of the triangle and $y$ being the height (due to the right angle), then $P=x+y+z$ and $A=frac{1}{2}xy$



                Then you can quote the formulae for primitive Pythagorean Triples, where $z=left(a^2+b^2right)$, $x=left(a^2-b^2right)$ and $y=2ab$



                To expand to all the non primitive Pythagorean triangles we have



                $$(cz)^2=(cx)^2+(cy)^2$$



                Therefore



                $$P=a(2a+2b)c$$
                $$A=frac{1}{2} left(a^2-b^2right)(2ab)c^2=left(a^2-b^2right)abc^2$$



                For two incongruent Pythagorean triangles 1 and 2 the condition is



                $$P_1=P_2 ;;text{and} ;; A_1=A_2$$



                For Perimeter:
                $$a_1(2a_1+2b_1)c_1=a_2(2a_2+2b_2)c_2$$
                $$frac{a_2+b_2}{a_1+b_1}=frac{a_1c_1}{a_2c_2}tag{1}$$
                For Area:
                $$left(a_1^2-b_1^2right)a_1b_1c_1^2=left(a_2^2-b_2^2right)a_2b_2c_2^2$$
                $$frac{a_2+b_2}{a_1+b_1}=frac{a_1c_1}{a_2c_2}left(frac{b_1c_1}{b_2c_2}frac{ (a_1-b_1) }{ (a_2-b_2) } right)$$



                Therefore combining both gives



                $$b_1c_1(a_1-b_1)=b_2c_2(a_2-b_2)tag{2}$$



                Update:



                Using (1) and (2) we can eliminate the variables $c_1$ and $c_2$ eventually giving
                $$frac{ (a_2+b_2) a_2 }{ (a_1+b_1) a_1 }=frac{ (a_2-b_2) b_2 }{ (a_1-b_1) b_1 }$$
                or
                $$frac{ (a_2+b_2) a_2 }{ (a_2-b_2) b_2 }=frac{ (a_1+b_1) a_1 }{ (a_1-b_1) b_1 }$$






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                • $begingroup$
                  In the version you use for primitive triples I think one needs $a,b$ odd and coprime, $a>b$ to give triples of positives. [That seems it would be known in that version…] I used the other version in my attempt, $p^2-q^2,2pq,p^2+q^2$ with $p,q$ coprime opposite parity and $p>q.$ Still didn't go to a finish in my attempts though.
                  $endgroup$
                  – coffeemath
                  18 hours ago






                • 1




                  $begingroup$
                  updated to your version
                  $endgroup$
                  – James Arathoon
                  17 hours ago












                Your Answer





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                5 Answers
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                7












                $begingroup$

                Consider two right triangles, with hypotenuses $p$ and $q$ and respective acute angles $theta$ and $phi$. To see that having equal perimeter and area makes them congruent, it suffices to show that
                $$theta = phi qquadtext{or}qquad theta+phi=frac{pi}{2} tag{0}$$
                (Either makes the triangles similar, which in turn makes them congruent.)



                Equating perimeters and areas gives a system we can write as



                $$begin{align}
                p(1+sintheta+costheta) &= q(1+sinphi+cosphi) \
                p^2 sintheta costheta &= q^2 sinphi cosphi
                end{align} tag{1}$$



                Defining $u:=tan(theta/2)$, we "know" that
                $$sintheta = frac{2u}{1+u^2} qquad costheta=frac{1-u^2}{1+u^2} quadtoquad 1+costheta+sintheta= frac{2 (1 + u)}{1 + u^2}$$
                and likewise for $v:=tan(phi/2)$. Thus, $(1)$ can be rewritten as
                $$begin{align}
                pfrac{(1+u)}{1+u^2} &= qfrac{(1+v)}{1+v^2} \[4pt]
                p^2frac{u(1+u)(1-u)}{(1+u^2)^2} &= q^2frac{v(1+v)(1-v)}{(1+v^2)^2}
                end{align}tag{2}$$

                Dividing the second equation by the square of the first ...



                $$frac{u(1-u)}{1+u} = frac{v(1-v)}{1+v} quadtoquad (u-v)(uv+u+v-1)=0 quadtoquad u=v, text{ or } frac{u+v}{1-uv}=1 tag{3}$$



                Therefore, we have one of the following situations (bearing in mind that $theta/2$ and $phi/2$ are each at most $pi/4$, so that we may draw appropriate conclusions from these tangent inequalities):
                $$begin{align}
                tanfrac{theta}{2}=tanfrac{phi}{2} &quadtoquad theta=phi \[4pt]
                frac{tan(theta/2)+tan(phi/2)}{1-tan(theta/2)tan(phi/2)} = 1 &quadtoquad
                tanleft(frac{theta}{2}+frac{phi}{2}right)=tanfrac{pi}{4} quadtoquad theta+phi=frac{pi}{2}end{align} tag{4}$$

                which match the sufficient conditions in $(0)$. $square$






                share|cite|improve this answer









                $endgroup$


















                  7












                  $begingroup$

                  Consider two right triangles, with hypotenuses $p$ and $q$ and respective acute angles $theta$ and $phi$. To see that having equal perimeter and area makes them congruent, it suffices to show that
                  $$theta = phi qquadtext{or}qquad theta+phi=frac{pi}{2} tag{0}$$
                  (Either makes the triangles similar, which in turn makes them congruent.)



                  Equating perimeters and areas gives a system we can write as



                  $$begin{align}
                  p(1+sintheta+costheta) &= q(1+sinphi+cosphi) \
                  p^2 sintheta costheta &= q^2 sinphi cosphi
                  end{align} tag{1}$$



                  Defining $u:=tan(theta/2)$, we "know" that
                  $$sintheta = frac{2u}{1+u^2} qquad costheta=frac{1-u^2}{1+u^2} quadtoquad 1+costheta+sintheta= frac{2 (1 + u)}{1 + u^2}$$
                  and likewise for $v:=tan(phi/2)$. Thus, $(1)$ can be rewritten as
                  $$begin{align}
                  pfrac{(1+u)}{1+u^2} &= qfrac{(1+v)}{1+v^2} \[4pt]
                  p^2frac{u(1+u)(1-u)}{(1+u^2)^2} &= q^2frac{v(1+v)(1-v)}{(1+v^2)^2}
                  end{align}tag{2}$$

                  Dividing the second equation by the square of the first ...



                  $$frac{u(1-u)}{1+u} = frac{v(1-v)}{1+v} quadtoquad (u-v)(uv+u+v-1)=0 quadtoquad u=v, text{ or } frac{u+v}{1-uv}=1 tag{3}$$



                  Therefore, we have one of the following situations (bearing in mind that $theta/2$ and $phi/2$ are each at most $pi/4$, so that we may draw appropriate conclusions from these tangent inequalities):
                  $$begin{align}
                  tanfrac{theta}{2}=tanfrac{phi}{2} &quadtoquad theta=phi \[4pt]
                  frac{tan(theta/2)+tan(phi/2)}{1-tan(theta/2)tan(phi/2)} = 1 &quadtoquad
                  tanleft(frac{theta}{2}+frac{phi}{2}right)=tanfrac{pi}{4} quadtoquad theta+phi=frac{pi}{2}end{align} tag{4}$$

                  which match the sufficient conditions in $(0)$. $square$






                  share|cite|improve this answer









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                    7












                    7








                    7





                    $begingroup$

                    Consider two right triangles, with hypotenuses $p$ and $q$ and respective acute angles $theta$ and $phi$. To see that having equal perimeter and area makes them congruent, it suffices to show that
                    $$theta = phi qquadtext{or}qquad theta+phi=frac{pi}{2} tag{0}$$
                    (Either makes the triangles similar, which in turn makes them congruent.)



                    Equating perimeters and areas gives a system we can write as



                    $$begin{align}
                    p(1+sintheta+costheta) &= q(1+sinphi+cosphi) \
                    p^2 sintheta costheta &= q^2 sinphi cosphi
                    end{align} tag{1}$$



                    Defining $u:=tan(theta/2)$, we "know" that
                    $$sintheta = frac{2u}{1+u^2} qquad costheta=frac{1-u^2}{1+u^2} quadtoquad 1+costheta+sintheta= frac{2 (1 + u)}{1 + u^2}$$
                    and likewise for $v:=tan(phi/2)$. Thus, $(1)$ can be rewritten as
                    $$begin{align}
                    pfrac{(1+u)}{1+u^2} &= qfrac{(1+v)}{1+v^2} \[4pt]
                    p^2frac{u(1+u)(1-u)}{(1+u^2)^2} &= q^2frac{v(1+v)(1-v)}{(1+v^2)^2}
                    end{align}tag{2}$$

                    Dividing the second equation by the square of the first ...



                    $$frac{u(1-u)}{1+u} = frac{v(1-v)}{1+v} quadtoquad (u-v)(uv+u+v-1)=0 quadtoquad u=v, text{ or } frac{u+v}{1-uv}=1 tag{3}$$



                    Therefore, we have one of the following situations (bearing in mind that $theta/2$ and $phi/2$ are each at most $pi/4$, so that we may draw appropriate conclusions from these tangent inequalities):
                    $$begin{align}
                    tanfrac{theta}{2}=tanfrac{phi}{2} &quadtoquad theta=phi \[4pt]
                    frac{tan(theta/2)+tan(phi/2)}{1-tan(theta/2)tan(phi/2)} = 1 &quadtoquad
                    tanleft(frac{theta}{2}+frac{phi}{2}right)=tanfrac{pi}{4} quadtoquad theta+phi=frac{pi}{2}end{align} tag{4}$$

                    which match the sufficient conditions in $(0)$. $square$






                    share|cite|improve this answer









                    $endgroup$



                    Consider two right triangles, with hypotenuses $p$ and $q$ and respective acute angles $theta$ and $phi$. To see that having equal perimeter and area makes them congruent, it suffices to show that
                    $$theta = phi qquadtext{or}qquad theta+phi=frac{pi}{2} tag{0}$$
                    (Either makes the triangles similar, which in turn makes them congruent.)



                    Equating perimeters and areas gives a system we can write as



                    $$begin{align}
                    p(1+sintheta+costheta) &= q(1+sinphi+cosphi) \
                    p^2 sintheta costheta &= q^2 sinphi cosphi
                    end{align} tag{1}$$



                    Defining $u:=tan(theta/2)$, we "know" that
                    $$sintheta = frac{2u}{1+u^2} qquad costheta=frac{1-u^2}{1+u^2} quadtoquad 1+costheta+sintheta= frac{2 (1 + u)}{1 + u^2}$$
                    and likewise for $v:=tan(phi/2)$. Thus, $(1)$ can be rewritten as
                    $$begin{align}
                    pfrac{(1+u)}{1+u^2} &= qfrac{(1+v)}{1+v^2} \[4pt]
                    p^2frac{u(1+u)(1-u)}{(1+u^2)^2} &= q^2frac{v(1+v)(1-v)}{(1+v^2)^2}
                    end{align}tag{2}$$

                    Dividing the second equation by the square of the first ...



                    $$frac{u(1-u)}{1+u} = frac{v(1-v)}{1+v} quadtoquad (u-v)(uv+u+v-1)=0 quadtoquad u=v, text{ or } frac{u+v}{1-uv}=1 tag{3}$$



                    Therefore, we have one of the following situations (bearing in mind that $theta/2$ and $phi/2$ are each at most $pi/4$, so that we may draw appropriate conclusions from these tangent inequalities):
                    $$begin{align}
                    tanfrac{theta}{2}=tanfrac{phi}{2} &quadtoquad theta=phi \[4pt]
                    frac{tan(theta/2)+tan(phi/2)}{1-tan(theta/2)tan(phi/2)} = 1 &quadtoquad
                    tanleft(frac{theta}{2}+frac{phi}{2}right)=tanfrac{pi}{4} quadtoquad theta+phi=frac{pi}{2}end{align} tag{4}$$

                    which match the sufficient conditions in $(0)$. $square$







                    share|cite|improve this answer












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                    share|cite|improve this answer










                    answered 15 hours ago









                    BlueBlue

                    49.6k870158




                    49.6k870158























                        8












                        $begingroup$

                        Let $S$ be the circumference and $A$ twice the area of a triangle.



                        Then,
                        $$a_i+b_i+sqrt{a_i^2+b_i^2}=S
                        text{ and } a_ib_i=A. tag{1}$$

                        After squaring, $S^2+2A-2S(a_i+b_i)=0$ and from here $a_i+b_i=S/2+A/S$. Thus, $c_i=S-(a_i+b_i)=S/2-A/S=c$, that is, triangles have the same hypotenuse.



                        Then, $a_i+b_i=S-c=T$ and $a_ib_i=A$, which results in a solutions for $a_i$ and $b_i$ expressed in terms of constants $A$ and $T$. Although one of the resulting equations is quadratic, there is a symmetric pair of solutions (thank you for comments below). Hence, all sides must be the same.






                        share|cite|improve this answer











                        $endgroup$









                        • 3




                          $begingroup$
                          Really nice proof. I have a niggle with your last statement though. In general, $2$ equations with $2$ unknowns may well have multiple solutions if they are not linear. For example, the OP mentions isosceles triangles, where the triangles $(8,8,12)$ and $(6,11,11)$ have the same perimeter and area even though isosceles triangles are also parametrised by two variables. Of course, in your case you know the sum and product of $a$ and $b$, which is a particularly nice pair of equations well known to have a symmetric pair of solutions.
                          $endgroup$
                          – Jaap Scherphuis
                          17 hours ago












                        • $begingroup$
                          @Jaap Scherphuis -- Thank you for your comment and clarification.
                          $endgroup$
                          – dnqxt
                          17 hours ago












                        • $begingroup$
                          @Dawood ibn Kareem -- Thanks for the comment. I added in the answer the clarification made by Jaap Scherphuis.
                          $endgroup$
                          – dnqxt
                          10 hours ago
















                        8












                        $begingroup$

                        Let $S$ be the circumference and $A$ twice the area of a triangle.



                        Then,
                        $$a_i+b_i+sqrt{a_i^2+b_i^2}=S
                        text{ and } a_ib_i=A. tag{1}$$

                        After squaring, $S^2+2A-2S(a_i+b_i)=0$ and from here $a_i+b_i=S/2+A/S$. Thus, $c_i=S-(a_i+b_i)=S/2-A/S=c$, that is, triangles have the same hypotenuse.



                        Then, $a_i+b_i=S-c=T$ and $a_ib_i=A$, which results in a solutions for $a_i$ and $b_i$ expressed in terms of constants $A$ and $T$. Although one of the resulting equations is quadratic, there is a symmetric pair of solutions (thank you for comments below). Hence, all sides must be the same.






                        share|cite|improve this answer











                        $endgroup$









                        • 3




                          $begingroup$
                          Really nice proof. I have a niggle with your last statement though. In general, $2$ equations with $2$ unknowns may well have multiple solutions if they are not linear. For example, the OP mentions isosceles triangles, where the triangles $(8,8,12)$ and $(6,11,11)$ have the same perimeter and area even though isosceles triangles are also parametrised by two variables. Of course, in your case you know the sum and product of $a$ and $b$, which is a particularly nice pair of equations well known to have a symmetric pair of solutions.
                          $endgroup$
                          – Jaap Scherphuis
                          17 hours ago












                        • $begingroup$
                          @Jaap Scherphuis -- Thank you for your comment and clarification.
                          $endgroup$
                          – dnqxt
                          17 hours ago












                        • $begingroup$
                          @Dawood ibn Kareem -- Thanks for the comment. I added in the answer the clarification made by Jaap Scherphuis.
                          $endgroup$
                          – dnqxt
                          10 hours ago














                        8












                        8








                        8





                        $begingroup$

                        Let $S$ be the circumference and $A$ twice the area of a triangle.



                        Then,
                        $$a_i+b_i+sqrt{a_i^2+b_i^2}=S
                        text{ and } a_ib_i=A. tag{1}$$

                        After squaring, $S^2+2A-2S(a_i+b_i)=0$ and from here $a_i+b_i=S/2+A/S$. Thus, $c_i=S-(a_i+b_i)=S/2-A/S=c$, that is, triangles have the same hypotenuse.



                        Then, $a_i+b_i=S-c=T$ and $a_ib_i=A$, which results in a solutions for $a_i$ and $b_i$ expressed in terms of constants $A$ and $T$. Although one of the resulting equations is quadratic, there is a symmetric pair of solutions (thank you for comments below). Hence, all sides must be the same.






                        share|cite|improve this answer











                        $endgroup$



                        Let $S$ be the circumference and $A$ twice the area of a triangle.



                        Then,
                        $$a_i+b_i+sqrt{a_i^2+b_i^2}=S
                        text{ and } a_ib_i=A. tag{1}$$

                        After squaring, $S^2+2A-2S(a_i+b_i)=0$ and from here $a_i+b_i=S/2+A/S$. Thus, $c_i=S-(a_i+b_i)=S/2-A/S=c$, that is, triangles have the same hypotenuse.



                        Then, $a_i+b_i=S-c=T$ and $a_ib_i=A$, which results in a solutions for $a_i$ and $b_i$ expressed in terms of constants $A$ and $T$. Although one of the resulting equations is quadratic, there is a symmetric pair of solutions (thank you for comments below). Hence, all sides must be the same.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited 10 hours ago

























                        answered 18 hours ago









                        dnqxtdnqxt

                        7075




                        7075








                        • 3




                          $begingroup$
                          Really nice proof. I have a niggle with your last statement though. In general, $2$ equations with $2$ unknowns may well have multiple solutions if they are not linear. For example, the OP mentions isosceles triangles, where the triangles $(8,8,12)$ and $(6,11,11)$ have the same perimeter and area even though isosceles triangles are also parametrised by two variables. Of course, in your case you know the sum and product of $a$ and $b$, which is a particularly nice pair of equations well known to have a symmetric pair of solutions.
                          $endgroup$
                          – Jaap Scherphuis
                          17 hours ago












                        • $begingroup$
                          @Jaap Scherphuis -- Thank you for your comment and clarification.
                          $endgroup$
                          – dnqxt
                          17 hours ago












                        • $begingroup$
                          @Dawood ibn Kareem -- Thanks for the comment. I added in the answer the clarification made by Jaap Scherphuis.
                          $endgroup$
                          – dnqxt
                          10 hours ago














                        • 3




                          $begingroup$
                          Really nice proof. I have a niggle with your last statement though. In general, $2$ equations with $2$ unknowns may well have multiple solutions if they are not linear. For example, the OP mentions isosceles triangles, where the triangles $(8,8,12)$ and $(6,11,11)$ have the same perimeter and area even though isosceles triangles are also parametrised by two variables. Of course, in your case you know the sum and product of $a$ and $b$, which is a particularly nice pair of equations well known to have a symmetric pair of solutions.
                          $endgroup$
                          – Jaap Scherphuis
                          17 hours ago












                        • $begingroup$
                          @Jaap Scherphuis -- Thank you for your comment and clarification.
                          $endgroup$
                          – dnqxt
                          17 hours ago












                        • $begingroup$
                          @Dawood ibn Kareem -- Thanks for the comment. I added in the answer the clarification made by Jaap Scherphuis.
                          $endgroup$
                          – dnqxt
                          10 hours ago








                        3




                        3




                        $begingroup$
                        Really nice proof. I have a niggle with your last statement though. In general, $2$ equations with $2$ unknowns may well have multiple solutions if they are not linear. For example, the OP mentions isosceles triangles, where the triangles $(8,8,12)$ and $(6,11,11)$ have the same perimeter and area even though isosceles triangles are also parametrised by two variables. Of course, in your case you know the sum and product of $a$ and $b$, which is a particularly nice pair of equations well known to have a symmetric pair of solutions.
                        $endgroup$
                        – Jaap Scherphuis
                        17 hours ago






                        $begingroup$
                        Really nice proof. I have a niggle with your last statement though. In general, $2$ equations with $2$ unknowns may well have multiple solutions if they are not linear. For example, the OP mentions isosceles triangles, where the triangles $(8,8,12)$ and $(6,11,11)$ have the same perimeter and area even though isosceles triangles are also parametrised by two variables. Of course, in your case you know the sum and product of $a$ and $b$, which is a particularly nice pair of equations well known to have a symmetric pair of solutions.
                        $endgroup$
                        – Jaap Scherphuis
                        17 hours ago














                        $begingroup$
                        @Jaap Scherphuis -- Thank you for your comment and clarification.
                        $endgroup$
                        – dnqxt
                        17 hours ago






                        $begingroup$
                        @Jaap Scherphuis -- Thank you for your comment and clarification.
                        $endgroup$
                        – dnqxt
                        17 hours ago














                        $begingroup$
                        @Dawood ibn Kareem -- Thanks for the comment. I added in the answer the clarification made by Jaap Scherphuis.
                        $endgroup$
                        – dnqxt
                        10 hours ago




                        $begingroup$
                        @Dawood ibn Kareem -- Thanks for the comment. I added in the answer the clarification made by Jaap Scherphuis.
                        $endgroup$
                        – dnqxt
                        10 hours ago











                        6












                        $begingroup$

                        Consider a right-angled triangle with sides $a$ and $b$.

                        The hypotenuse has length $c=sqrt{a^2+b^2}$.
                        Its area is $ab/2$, and perimeter is $a+b+c$.



                        I will allow $a,b,c$ to be any positive real numbers, not restrict them to positive integers.



                        Suppose we scale the triangles such that $ab=1$, (i.e. an area of $1/2$). Is it possible to have two of these triangles that are distinct but with the same perimeter?



                        We may assume that $a$ is the longer side, i.e. $a>b$, so we must have $a>1$.



                        The perimeter is
                        $$P(a) = a+b+c\ = a+b+sqrt{a^2+b^2}\ = a+frac{1}{a}+sqrt{a^2+frac{1}{a^2}}$$



                        This is an increasing function on the interval $[1,infty)$ because its derivative w.r.t. $a$ is positive for $a>1$. This is tedious to check by hand, so I used Wolfram alpha. You can however understand why this is the case by noticing that if you increase $a$, then the rate at which $a$ increases is larger than the rate at which $1/a$ decreases, and the same holds for $a^2$ versus $1/a^2$.



                        This means that there are no two values of $a$, both with $a>1$ for which you get the same perimeter.



                        Bringing it back to the original problem, it means that there are no two right-angled triangles with the same perimeter and area, unless they have the same sides. Basically, given an area and a perimeter, their two equations uniquely determine the two triangle sides because the lines those equations represent are not curved enough to intersect multiple times.





                        P.S. By the way, the OP mentioned that there are pairs of isosceles triangles with matching areas and perimeters, even when all sides and the area are integers. Two examples are:
                        $(29,29,40)$ and $(37,37,24)$,

                        and also
                        $(218,218,240)$ and $(233,233,210)$.






                        share|cite|improve this answer











                        $endgroup$


















                          6












                          $begingroup$

                          Consider a right-angled triangle with sides $a$ and $b$.

                          The hypotenuse has length $c=sqrt{a^2+b^2}$.
                          Its area is $ab/2$, and perimeter is $a+b+c$.



                          I will allow $a,b,c$ to be any positive real numbers, not restrict them to positive integers.



                          Suppose we scale the triangles such that $ab=1$, (i.e. an area of $1/2$). Is it possible to have two of these triangles that are distinct but with the same perimeter?



                          We may assume that $a$ is the longer side, i.e. $a>b$, so we must have $a>1$.



                          The perimeter is
                          $$P(a) = a+b+c\ = a+b+sqrt{a^2+b^2}\ = a+frac{1}{a}+sqrt{a^2+frac{1}{a^2}}$$



                          This is an increasing function on the interval $[1,infty)$ because its derivative w.r.t. $a$ is positive for $a>1$. This is tedious to check by hand, so I used Wolfram alpha. You can however understand why this is the case by noticing that if you increase $a$, then the rate at which $a$ increases is larger than the rate at which $1/a$ decreases, and the same holds for $a^2$ versus $1/a^2$.



                          This means that there are no two values of $a$, both with $a>1$ for which you get the same perimeter.



                          Bringing it back to the original problem, it means that there are no two right-angled triangles with the same perimeter and area, unless they have the same sides. Basically, given an area and a perimeter, their two equations uniquely determine the two triangle sides because the lines those equations represent are not curved enough to intersect multiple times.





                          P.S. By the way, the OP mentioned that there are pairs of isosceles triangles with matching areas and perimeters, even when all sides and the area are integers. Two examples are:
                          $(29,29,40)$ and $(37,37,24)$,

                          and also
                          $(218,218,240)$ and $(233,233,210)$.






                          share|cite|improve this answer











                          $endgroup$
















                            6












                            6








                            6





                            $begingroup$

                            Consider a right-angled triangle with sides $a$ and $b$.

                            The hypotenuse has length $c=sqrt{a^2+b^2}$.
                            Its area is $ab/2$, and perimeter is $a+b+c$.



                            I will allow $a,b,c$ to be any positive real numbers, not restrict them to positive integers.



                            Suppose we scale the triangles such that $ab=1$, (i.e. an area of $1/2$). Is it possible to have two of these triangles that are distinct but with the same perimeter?



                            We may assume that $a$ is the longer side, i.e. $a>b$, so we must have $a>1$.



                            The perimeter is
                            $$P(a) = a+b+c\ = a+b+sqrt{a^2+b^2}\ = a+frac{1}{a}+sqrt{a^2+frac{1}{a^2}}$$



                            This is an increasing function on the interval $[1,infty)$ because its derivative w.r.t. $a$ is positive for $a>1$. This is tedious to check by hand, so I used Wolfram alpha. You can however understand why this is the case by noticing that if you increase $a$, then the rate at which $a$ increases is larger than the rate at which $1/a$ decreases, and the same holds for $a^2$ versus $1/a^2$.



                            This means that there are no two values of $a$, both with $a>1$ for which you get the same perimeter.



                            Bringing it back to the original problem, it means that there are no two right-angled triangles with the same perimeter and area, unless they have the same sides. Basically, given an area and a perimeter, their two equations uniquely determine the two triangle sides because the lines those equations represent are not curved enough to intersect multiple times.





                            P.S. By the way, the OP mentioned that there are pairs of isosceles triangles with matching areas and perimeters, even when all sides and the area are integers. Two examples are:
                            $(29,29,40)$ and $(37,37,24)$,

                            and also
                            $(218,218,240)$ and $(233,233,210)$.






                            share|cite|improve this answer











                            $endgroup$



                            Consider a right-angled triangle with sides $a$ and $b$.

                            The hypotenuse has length $c=sqrt{a^2+b^2}$.
                            Its area is $ab/2$, and perimeter is $a+b+c$.



                            I will allow $a,b,c$ to be any positive real numbers, not restrict them to positive integers.



                            Suppose we scale the triangles such that $ab=1$, (i.e. an area of $1/2$). Is it possible to have two of these triangles that are distinct but with the same perimeter?



                            We may assume that $a$ is the longer side, i.e. $a>b$, so we must have $a>1$.



                            The perimeter is
                            $$P(a) = a+b+c\ = a+b+sqrt{a^2+b^2}\ = a+frac{1}{a}+sqrt{a^2+frac{1}{a^2}}$$



                            This is an increasing function on the interval $[1,infty)$ because its derivative w.r.t. $a$ is positive for $a>1$. This is tedious to check by hand, so I used Wolfram alpha. You can however understand why this is the case by noticing that if you increase $a$, then the rate at which $a$ increases is larger than the rate at which $1/a$ decreases, and the same holds for $a^2$ versus $1/a^2$.



                            This means that there are no two values of $a$, both with $a>1$ for which you get the same perimeter.



                            Bringing it back to the original problem, it means that there are no two right-angled triangles with the same perimeter and area, unless they have the same sides. Basically, given an area and a perimeter, their two equations uniquely determine the two triangle sides because the lines those equations represent are not curved enough to intersect multiple times.





                            P.S. By the way, the OP mentioned that there are pairs of isosceles triangles with matching areas and perimeters, even when all sides and the area are integers. Two examples are:
                            $(29,29,40)$ and $(37,37,24)$,

                            and also
                            $(218,218,240)$ and $(233,233,210)$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 15 hours ago

























                            answered 17 hours ago









                            Jaap ScherphuisJaap Scherphuis

                            4,347817




                            4,347817























                                5












                                $begingroup$

                                The area and the perimeter uniquely define the radius of the inscribed circle because $A=frac12Pr$, and the hypotenuse because $r=frac{P}{2}-c$. That fixes both $a+b=P-c$ and $ab=2A$ so $a$ and $b$ are also unique up to permutation, QED.






                                share|cite|improve this answer











                                $endgroup$


















                                  5












                                  $begingroup$

                                  The area and the perimeter uniquely define the radius of the inscribed circle because $A=frac12Pr$, and the hypotenuse because $r=frac{P}{2}-c$. That fixes both $a+b=P-c$ and $ab=2A$ so $a$ and $b$ are also unique up to permutation, QED.






                                  share|cite|improve this answer











                                  $endgroup$
















                                    5












                                    5








                                    5





                                    $begingroup$

                                    The area and the perimeter uniquely define the radius of the inscribed circle because $A=frac12Pr$, and the hypotenuse because $r=frac{P}{2}-c$. That fixes both $a+b=P-c$ and $ab=2A$ so $a$ and $b$ are also unique up to permutation, QED.






                                    share|cite|improve this answer











                                    $endgroup$



                                    The area and the perimeter uniquely define the radius of the inscribed circle because $A=frac12Pr$, and the hypotenuse because $r=frac{P}{2}-c$. That fixes both $a+b=P-c$ and $ab=2A$ so $a$ and $b$ are also unique up to permutation, QED.







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited 11 hours ago

























                                    answered 11 hours ago









                                    Roman OdaiskyRoman Odaisky

                                    24116




                                    24116























                                        2












                                        $begingroup$

                                        Long Comment:



                                        You can at least give the simple formulae for the perimeter $P$ and area $A$ of a right angle triangle.



                                        If $z^2=x^2+y^2$ is a primitive Pythagorean triangle with $x$ being the base of the triangle and $y$ being the height (due to the right angle), then $P=x+y+z$ and $A=frac{1}{2}xy$



                                        Then you can quote the formulae for primitive Pythagorean Triples, where $z=left(a^2+b^2right)$, $x=left(a^2-b^2right)$ and $y=2ab$



                                        To expand to all the non primitive Pythagorean triangles we have



                                        $$(cz)^2=(cx)^2+(cy)^2$$



                                        Therefore



                                        $$P=a(2a+2b)c$$
                                        $$A=frac{1}{2} left(a^2-b^2right)(2ab)c^2=left(a^2-b^2right)abc^2$$



                                        For two incongruent Pythagorean triangles 1 and 2 the condition is



                                        $$P_1=P_2 ;;text{and} ;; A_1=A_2$$



                                        For Perimeter:
                                        $$a_1(2a_1+2b_1)c_1=a_2(2a_2+2b_2)c_2$$
                                        $$frac{a_2+b_2}{a_1+b_1}=frac{a_1c_1}{a_2c_2}tag{1}$$
                                        For Area:
                                        $$left(a_1^2-b_1^2right)a_1b_1c_1^2=left(a_2^2-b_2^2right)a_2b_2c_2^2$$
                                        $$frac{a_2+b_2}{a_1+b_1}=frac{a_1c_1}{a_2c_2}left(frac{b_1c_1}{b_2c_2}frac{ (a_1-b_1) }{ (a_2-b_2) } right)$$



                                        Therefore combining both gives



                                        $$b_1c_1(a_1-b_1)=b_2c_2(a_2-b_2)tag{2}$$



                                        Update:



                                        Using (1) and (2) we can eliminate the variables $c_1$ and $c_2$ eventually giving
                                        $$frac{ (a_2+b_2) a_2 }{ (a_1+b_1) a_1 }=frac{ (a_2-b_2) b_2 }{ (a_1-b_1) b_1 }$$
                                        or
                                        $$frac{ (a_2+b_2) a_2 }{ (a_2-b_2) b_2 }=frac{ (a_1+b_1) a_1 }{ (a_1-b_1) b_1 }$$






                                        share|cite|improve this answer











                                        $endgroup$













                                        • $begingroup$
                                          In the version you use for primitive triples I think one needs $a,b$ odd and coprime, $a>b$ to give triples of positives. [That seems it would be known in that version…] I used the other version in my attempt, $p^2-q^2,2pq,p^2+q^2$ with $p,q$ coprime opposite parity and $p>q.$ Still didn't go to a finish in my attempts though.
                                          $endgroup$
                                          – coffeemath
                                          18 hours ago






                                        • 1




                                          $begingroup$
                                          updated to your version
                                          $endgroup$
                                          – James Arathoon
                                          17 hours ago
















                                        2












                                        $begingroup$

                                        Long Comment:



                                        You can at least give the simple formulae for the perimeter $P$ and area $A$ of a right angle triangle.



                                        If $z^2=x^2+y^2$ is a primitive Pythagorean triangle with $x$ being the base of the triangle and $y$ being the height (due to the right angle), then $P=x+y+z$ and $A=frac{1}{2}xy$



                                        Then you can quote the formulae for primitive Pythagorean Triples, where $z=left(a^2+b^2right)$, $x=left(a^2-b^2right)$ and $y=2ab$



                                        To expand to all the non primitive Pythagorean triangles we have



                                        $$(cz)^2=(cx)^2+(cy)^2$$



                                        Therefore



                                        $$P=a(2a+2b)c$$
                                        $$A=frac{1}{2} left(a^2-b^2right)(2ab)c^2=left(a^2-b^2right)abc^2$$



                                        For two incongruent Pythagorean triangles 1 and 2 the condition is



                                        $$P_1=P_2 ;;text{and} ;; A_1=A_2$$



                                        For Perimeter:
                                        $$a_1(2a_1+2b_1)c_1=a_2(2a_2+2b_2)c_2$$
                                        $$frac{a_2+b_2}{a_1+b_1}=frac{a_1c_1}{a_2c_2}tag{1}$$
                                        For Area:
                                        $$left(a_1^2-b_1^2right)a_1b_1c_1^2=left(a_2^2-b_2^2right)a_2b_2c_2^2$$
                                        $$frac{a_2+b_2}{a_1+b_1}=frac{a_1c_1}{a_2c_2}left(frac{b_1c_1}{b_2c_2}frac{ (a_1-b_1) }{ (a_2-b_2) } right)$$



                                        Therefore combining both gives



                                        $$b_1c_1(a_1-b_1)=b_2c_2(a_2-b_2)tag{2}$$



                                        Update:



                                        Using (1) and (2) we can eliminate the variables $c_1$ and $c_2$ eventually giving
                                        $$frac{ (a_2+b_2) a_2 }{ (a_1+b_1) a_1 }=frac{ (a_2-b_2) b_2 }{ (a_1-b_1) b_1 }$$
                                        or
                                        $$frac{ (a_2+b_2) a_2 }{ (a_2-b_2) b_2 }=frac{ (a_1+b_1) a_1 }{ (a_1-b_1) b_1 }$$






                                        share|cite|improve this answer











                                        $endgroup$













                                        • $begingroup$
                                          In the version you use for primitive triples I think one needs $a,b$ odd and coprime, $a>b$ to give triples of positives. [That seems it would be known in that version…] I used the other version in my attempt, $p^2-q^2,2pq,p^2+q^2$ with $p,q$ coprime opposite parity and $p>q.$ Still didn't go to a finish in my attempts though.
                                          $endgroup$
                                          – coffeemath
                                          18 hours ago






                                        • 1




                                          $begingroup$
                                          updated to your version
                                          $endgroup$
                                          – James Arathoon
                                          17 hours ago














                                        2












                                        2








                                        2





                                        $begingroup$

                                        Long Comment:



                                        You can at least give the simple formulae for the perimeter $P$ and area $A$ of a right angle triangle.



                                        If $z^2=x^2+y^2$ is a primitive Pythagorean triangle with $x$ being the base of the triangle and $y$ being the height (due to the right angle), then $P=x+y+z$ and $A=frac{1}{2}xy$



                                        Then you can quote the formulae for primitive Pythagorean Triples, where $z=left(a^2+b^2right)$, $x=left(a^2-b^2right)$ and $y=2ab$



                                        To expand to all the non primitive Pythagorean triangles we have



                                        $$(cz)^2=(cx)^2+(cy)^2$$



                                        Therefore



                                        $$P=a(2a+2b)c$$
                                        $$A=frac{1}{2} left(a^2-b^2right)(2ab)c^2=left(a^2-b^2right)abc^2$$



                                        For two incongruent Pythagorean triangles 1 and 2 the condition is



                                        $$P_1=P_2 ;;text{and} ;; A_1=A_2$$



                                        For Perimeter:
                                        $$a_1(2a_1+2b_1)c_1=a_2(2a_2+2b_2)c_2$$
                                        $$frac{a_2+b_2}{a_1+b_1}=frac{a_1c_1}{a_2c_2}tag{1}$$
                                        For Area:
                                        $$left(a_1^2-b_1^2right)a_1b_1c_1^2=left(a_2^2-b_2^2right)a_2b_2c_2^2$$
                                        $$frac{a_2+b_2}{a_1+b_1}=frac{a_1c_1}{a_2c_2}left(frac{b_1c_1}{b_2c_2}frac{ (a_1-b_1) }{ (a_2-b_2) } right)$$



                                        Therefore combining both gives



                                        $$b_1c_1(a_1-b_1)=b_2c_2(a_2-b_2)tag{2}$$



                                        Update:



                                        Using (1) and (2) we can eliminate the variables $c_1$ and $c_2$ eventually giving
                                        $$frac{ (a_2+b_2) a_2 }{ (a_1+b_1) a_1 }=frac{ (a_2-b_2) b_2 }{ (a_1-b_1) b_1 }$$
                                        or
                                        $$frac{ (a_2+b_2) a_2 }{ (a_2-b_2) b_2 }=frac{ (a_1+b_1) a_1 }{ (a_1-b_1) b_1 }$$






                                        share|cite|improve this answer











                                        $endgroup$



                                        Long Comment:



                                        You can at least give the simple formulae for the perimeter $P$ and area $A$ of a right angle triangle.



                                        If $z^2=x^2+y^2$ is a primitive Pythagorean triangle with $x$ being the base of the triangle and $y$ being the height (due to the right angle), then $P=x+y+z$ and $A=frac{1}{2}xy$



                                        Then you can quote the formulae for primitive Pythagorean Triples, where $z=left(a^2+b^2right)$, $x=left(a^2-b^2right)$ and $y=2ab$



                                        To expand to all the non primitive Pythagorean triangles we have



                                        $$(cz)^2=(cx)^2+(cy)^2$$



                                        Therefore



                                        $$P=a(2a+2b)c$$
                                        $$A=frac{1}{2} left(a^2-b^2right)(2ab)c^2=left(a^2-b^2right)abc^2$$



                                        For two incongruent Pythagorean triangles 1 and 2 the condition is



                                        $$P_1=P_2 ;;text{and} ;; A_1=A_2$$



                                        For Perimeter:
                                        $$a_1(2a_1+2b_1)c_1=a_2(2a_2+2b_2)c_2$$
                                        $$frac{a_2+b_2}{a_1+b_1}=frac{a_1c_1}{a_2c_2}tag{1}$$
                                        For Area:
                                        $$left(a_1^2-b_1^2right)a_1b_1c_1^2=left(a_2^2-b_2^2right)a_2b_2c_2^2$$
                                        $$frac{a_2+b_2}{a_1+b_1}=frac{a_1c_1}{a_2c_2}left(frac{b_1c_1}{b_2c_2}frac{ (a_1-b_1) }{ (a_2-b_2) } right)$$



                                        Therefore combining both gives



                                        $$b_1c_1(a_1-b_1)=b_2c_2(a_2-b_2)tag{2}$$



                                        Update:



                                        Using (1) and (2) we can eliminate the variables $c_1$ and $c_2$ eventually giving
                                        $$frac{ (a_2+b_2) a_2 }{ (a_1+b_1) a_1 }=frac{ (a_2-b_2) b_2 }{ (a_1-b_1) b_1 }$$
                                        or
                                        $$frac{ (a_2+b_2) a_2 }{ (a_2-b_2) b_2 }=frac{ (a_1+b_1) a_1 }{ (a_1-b_1) b_1 }$$







                                        share|cite|improve this answer














                                        share|cite|improve this answer



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                                        edited 10 hours ago

























                                        answered 18 hours ago









                                        James ArathoonJames Arathoon

                                        1,608423




                                        1,608423












                                        • $begingroup$
                                          In the version you use for primitive triples I think one needs $a,b$ odd and coprime, $a>b$ to give triples of positives. [That seems it would be known in that version…] I used the other version in my attempt, $p^2-q^2,2pq,p^2+q^2$ with $p,q$ coprime opposite parity and $p>q.$ Still didn't go to a finish in my attempts though.
                                          $endgroup$
                                          – coffeemath
                                          18 hours ago






                                        • 1




                                          $begingroup$
                                          updated to your version
                                          $endgroup$
                                          – James Arathoon
                                          17 hours ago


















                                        • $begingroup$
                                          In the version you use for primitive triples I think one needs $a,b$ odd and coprime, $a>b$ to give triples of positives. [That seems it would be known in that version…] I used the other version in my attempt, $p^2-q^2,2pq,p^2+q^2$ with $p,q$ coprime opposite parity and $p>q.$ Still didn't go to a finish in my attempts though.
                                          $endgroup$
                                          – coffeemath
                                          18 hours ago






                                        • 1




                                          $begingroup$
                                          updated to your version
                                          $endgroup$
                                          – James Arathoon
                                          17 hours ago
















                                        $begingroup$
                                        In the version you use for primitive triples I think one needs $a,b$ odd and coprime, $a>b$ to give triples of positives. [That seems it would be known in that version…] I used the other version in my attempt, $p^2-q^2,2pq,p^2+q^2$ with $p,q$ coprime opposite parity and $p>q.$ Still didn't go to a finish in my attempts though.
                                        $endgroup$
                                        – coffeemath
                                        18 hours ago




                                        $begingroup$
                                        In the version you use for primitive triples I think one needs $a,b$ odd and coprime, $a>b$ to give triples of positives. [That seems it would be known in that version…] I used the other version in my attempt, $p^2-q^2,2pq,p^2+q^2$ with $p,q$ coprime opposite parity and $p>q.$ Still didn't go to a finish in my attempts though.
                                        $endgroup$
                                        – coffeemath
                                        18 hours ago




                                        1




                                        1




                                        $begingroup$
                                        updated to your version
                                        $endgroup$
                                        – James Arathoon
                                        17 hours ago




                                        $begingroup$
                                        updated to your version
                                        $endgroup$
                                        – James Arathoon
                                        17 hours ago


















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