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?
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).
Let $n \in \mathbb{N}$ and $a_i, b_i, m \in \mathbb{Z}$ for $i=1,2,\dots,k$.
If
$$
\sum_{i=1}^k a_i^n = \sum_{i=1}^k b_i^n = m$$
then
$$
\left( \sum_{i=1}^k a_i^n \right) \left( \sum_{i=1}^k b_i^n \right) = m^2.
$$
This Result can be further Generalized to cubes and higher powers(product of $x$ number of Sums is equal to $m^x$) .
If $$m=a^n-b^n=c^n-d^n$$ for $n \in \{1,2,3,4\}$ and $a,b,c,d \in \mathbb{N}$ then $$(a^n-b^n)(c^n-d^n)=m^2$$
Parametrizations for $n=2$ :
$1$st parametrization :
Take $p=ac+bd$, $q=ac-bd$, $r=ad+bc$ and $s=ad-bc$
Wkt $p^2-q^2=r^2-s^2=4abcd$
Substitute it for above result then we get
$(p^2-q^2)(r^2-s^2)=(4abcd)^2$
$2$nd parametrization :
Take $p=ac+bd$ , $q=ad+bc$ , $r=ad-bc$ and $s=ac-bd$
Wkt $p^2-q^2=r^2-s^2=(a^2-b^2)(c^2-d^2)$
Substitute it for above result then we get
$(p^2-q^2)(r^2-s^2)=(a^2-b^2)^2(c^2-d^2)^2$
Similarly for $n=3$ and $n=4$ we can obtain parametrizations.
Some examples :
For $n=2$ :
Wkt $4^2+7^2=1^2+8^2$
$(4^2-1^2)(8^2-7^2)=15^2$
$(7^2-1^2)(8^2-1^2)=48^2$
For $n=3$ :
Wkt $10^3+9^3=1^3+12^3$
$(9^3-1^3)(12^3-10^3)=(6^3+8^3)^2=728^2$
$(10^3-1^3)(12^3-9^3)=999^2$
For $n=4$ :
Wkt $133^4+134^4=59^4+158^4$
$(133^4-59^4)(158^4-134^4)=300783360^2$
$(134^4-59^4)(158^4-133^4)=310300575^2$
