<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] We have the below Identity:
$(b^2+c^2-a^2)^2+(2ab)^2+(2ac)^2=(b^2+c^2+a^2)^2$
Let:
$p=(b^2+c^2-a^2)^2$
$q=(2ab)^2$
$r=(2ac)^2$
$p+q+r=(a^2+b^2+c^2)^2$
To make (RHS) a fourth power we take:
$b^2+c^2=8a^2$  ---$(1)$
Hence,
$p+q+r=(9a^2)^2=(3a)^4$
$(pqr)=((b^2+c^2-a^2)(2ab)(2ac))^2$
=$(2a)^4(bc(b^2+c^2-a^2))^2$
Since, $(b^2+c^2=8a^2)$ we have $(b^2+c^2-a^2)=(7a^2)$
and to make, $(bc(b^2+c^2-a^2))$, a square we take,
$b=7c$  ---$(2)$
Hence,
$pqr=(2a)^4(7ac)^4=(14a^2c)^4$
Now eqn (1) & (2) above is satisfied by $(a,b,c)=(5,14,2)$
Hence:
$(p+q+r)=(15)^4$
$(p,q,r)=(700)^4$
And,
$(p,q,r)=(30625,19600,400)$
</s>