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I met in many places the equation

$(a^4-b^4)(c^4-d^4)=\square$

It is well known that this was investigated by Euler. But I was unable to find the general solution of this equation. Could you please clarify if there is such solution and if so could you please help to find the solution?

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  • $\begingroup$ I may not be familiar with the field, but what does the ractangle sign mean in this context ?(for me it's a d'alembertian) $\endgroup$
    – Amir Sagiv
    Apr 30, 2016 at 9:08
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    $\begingroup$ @Amir I guess it is 'a perfect square' $\endgroup$ Apr 30, 2016 at 9:15
  • $\begingroup$ Yes, the square is perfect square. $\endgroup$
    – veg_nw
    Apr 30, 2016 at 10:52

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I think general solution doesn't exist via elliptic curves and probably this can be proven rigorously.

Pick random $c,d$ where $c \ne \pm d$ and let $k=c^4-d^4$.

So we get $k(a^4-b^4)=z^2$. Dividing by $b^4$ we get $C:k(a'^4-1)=z'^2$.

This is elliptic curve and if it is of positive rank, it has infinitely many rational solutions $a'=a''/b'',z'=z''/b''$ coming from the group law. Multiplying by $b''^4$ we get the integer solutions $k(a''^4-b''^4)=z''^2 b''^2$.

On $C$ setting $a'=c/d$ gives one rational point and if it is of infinite order, it gives infinitely many solutions.

A single $k$ with positive rank gives you infinitely many integer solutions.

I suspect this happens for infinitely many $k$.


The OP asks about general polynomial solution.

Single $k$ with positive rank will give for fixed $c,d$ infinitely many solutions $a,b$ which are arbitrary large, ruling out polynomial solution.

Another approach might be to examine the K3 surface $a^4-b^4=c^4-d^4$ which gives subset of the solutions and again doesn't have complete polynomial parametrization (essentially sum of two fourth powers in two nontrivial ways, for which at least two polynomial parametrization are known).

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  • $\begingroup$ If I understand correctly the general solution does not exist? If this can be proven does this mean that there is no polynominal solution of this equation in general? $\endgroup$
    – veg_nw
    Apr 30, 2016 at 10:58
  • $\begingroup$ @veg_nw Yes, I am ready to bet polynomial solution doesn't exist. $\endgroup$
    – joro
    Apr 30, 2016 at 11:07
  • $\begingroup$ @joro You mean apart from the silly stuff like $a=kc,b=kd$? $\endgroup$ Apr 30, 2016 at 11:09
  • $\begingroup$ @LevBorisov Yes. This is not complete parametrization. Though setting $a'=c/d$ gives a point on the elliptic curve and remains to show it is of infinite order. $\endgroup$
    – joro
    Apr 30, 2016 at 11:11
  • $\begingroup$ And does this mean that there is no general "non-polynominal" solutions exist as well? $\endgroup$
    – veg_nw
    Apr 30, 2016 at 12:14

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