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I am solving this Diophantine equation

$$xy(x^2-y^2)=pq(p^2-q^2)=rs(r^2-s^2)$$

where $(x^2+y^2)^2+(p^2+q^2)^2=2(r^2+s^2)^2$,

but I am not able to proceed.

I use Euler method $pq(p^2-q^2) = rs(r^2-s^2)$.

If $s = p$, the equation reduces to solving $p^2 = q^2-qr+r^2$, two solutions of which are,

$$\{p,q,r\}=\{u^2+3v^2, u^2+2uv-3v^2, 4uv\}$$

$$\{p,q,r\}=\{u^2+3v^2, u^2-2uv-3v^2,u^2+2uv-3v^2\}.$$

But L.J.Lander gives the solution

$$\{p,q,r,s\} = \{v^5-2v, v^5+v, -2v^4+1, v^4+1\}.$$

But I am to find integer solutions for the above three equivalent rather than two with specific conditions $$(x^2+y^2)^2+(p^2+q^2)^2=2(r^2+s^2)^2$$ which will help me to know more about arithmetic progression between square. So, please provide solution or guidance to solve this type of diophantine equation or write a computer program to get integer solutions.

M. A.
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Anurag
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1 Answers1

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For reference, the following results are known for the pythagorean triangles with same area. According to Sequence of Triplets of Primitive Pythagorean Triangles with Same Area(https://proofwiki.org/wiki/Sequence_of_Triplets_of_Primitive_Pythagorean_Triangles_with_Same_Area), integer solutions of $xy(x^2-y^2)=pq(p^2-q^2)=rs(r^2-s^2)$ are given below. But it doesn't satisfy the second equation.

                (x,y,p,q,r,s,area), area=xy(x^2-y^2)

            (77, 38, 78, 55, 138, 5, 13123110)
            (2035, 266, 3306, 61, 3422, 55, 2203385574390)
            (1610, 869, 2002, 1817, 2622, 143, 2570042985510)
            (2201, 1166, 2438, 2035, 3565, 198, 8943387723270)
            (7238, 2465, 9077, 1122, 10434, 731, 826290896699730)
            (352538, 2999447,1931103, 2398838,3063347, 3215070,9381843970167926138271390)

Addendum

The formulae for triplets of Pythagorean triangles with Same Area are known below. (Is there a formula for this specific pattern of Pythagorean Triangles sharing an area?) It seems that the two families of solutions have no integer solution for $(x^2+y^2)^2+(p^2+q^2)^2=2(r^2+s^2)^2$.

$M_1 N_1 (M_1^2-N_1^2)=M_2 N_2 (M_2^2-N_2^2)=M_3 N_3 (M_3^2-N_3^2)$

1st family (Enrique Zeleny). Special case: $M_1+N_1 = M_2+N_2 = M_3-N_3$

$M_1 = r^2 + rs + s^2;\quad N_1 = r^2 - s^2$

$M_2 = r^2 + rs + s^2;\quad N_2 = 2rs + s^2$

$M_3 = r^2 + 2rs;\quad\quad\; N_3 = r^2 + rs + s^2$

2nd family (Tito Piezas III). Special case: $M_1+N_1 = M_2+N_2 = M_3-N_3$

$M_1 = (u + v) (2u + 3 v);\quad N_1 = v (u + 3 v)$

$M_2 = (u + 2 v) (u + 3 v);\quad N_2 = u(u + v)$

$M_3 = (u + 2 v) (2u + 3 v);\quad N_3 = uv$

Tomita
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