<s>[INST] Solving the Diophantine system $pqr=a^4$, $p+q+r=b^4$ I am trying to find solutions of the following system of diophantine equations:
$$\left\{\begin{array}{rcl}pqr&=&a^4\\p+q+r&=&b^4\end{array}\right.$$
where $a$, $b$, $p$, $q$ ans $r$ are positive integers such that $\gcd(p,q,r)$ is not divisible by $\theta^4$, $\theta>1$.
I found the following solutions $(p,q,r)$ with a computer program :
$(3\,;6\,;72)$ , $(25\,;60\,;540)$ , $(72\,;576\,;648)$ and $(162\,;448\,;686)$.
The system has infinitely many solutions : take $(p\;q\,;r)=\left(A^4\,;B^4\,;C^4\right)$, where $A^4+B^4+C^4=D^4$ and $A$, $B$, $C$ are coprime (see this article).
But can we prove that there are infinitely many solutions using more elementary ways ?
Thank you for your help !
 [/INST] Take,
$2p=2n^2+1-w$
$2q=2n^2+1+w$
$2r=16m^2$
Where,
$w^2=4n^4+4n^2-8m^2+1$ ---$(1)$
Eqn (1), is satisfied at, $(m,n,w)=(3,2,3)$
Hence,
$p+q+r=8m^2+2n^2+1$
For, $(m,n,w)=(3,2,3)$
$p+q+r=(3)^4$
$8pqr=(2p)(2q)(2r)$
=$(2n^2+1-w)(2n^2+1+w)(16m^2)$
For, $(m,n,w)=(3,2,3)$, we get:
$pqr=(6)^4$
$(p,q,r)=(3,6,72)$

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