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 !
 A: Thanks to John Ca for alerting me to this question.
It is equivalent to finding infinitely many solutions to
the Diophantine equation $x y z (x+y+z) = 1$
in positive rational numbers $(x,y,z) = (p,q,r)/ab$.
That's a special case of a problem that was already solved by Euler in 1749,
though we do not know how he did it; see my lecture notes
https://abel.math.harvard.edu/~elkies/euler_14t.pdf
for some of the mathematical context.
His formula is on page 9; taking $a=1$
we find the following parametrization (I dehomogenized by setting s=1):
$$
x = \frac{6 t^3 (t^4-2)^2} {(4 t^4 + 1)   (2 t^8 + 10 t^4 - 1)},
$$ $$
y = \frac{ 3 (4 t^4 + 1)^2} {2t (t^4-2) (2 t^8 + 10 t^4 - 1)},
$$ $$
z = \frac{ 2 (2 t^8 + 10 t^4 - 1)} {3t (4 t^4 + 1)}.
$$
In computer-readable format, $x,y,z$ are
6 * t^3 * (t^4-2)^2 / ((4 * t^4+1) * (2 * t^8 + 10 * t^4 - 1)),
3 * (4 * t^4+1)^2 / (2 * t * (t^4-2) * (2 * t^8 + 10 * t^4 - 1)),
2 * (2 * t^8 + 10 * t^4 - 1) / (3 * t * (4 * t^4+1))

They are all positive for $t > 2^{1/4}$; Euler gives the example $t=2$
which makes $(x,y,z) = (9408/43615, 12675/37576, 671/195)$;
multiplying through by $1342^3 65^4 21$ gives the solution
$$
(p,q,r) = 65^3 671^2 (1580544, 2471625, 25213496)
$$
with $p+q+r = 43615^4$, $pqr = 20772769017000^4$,
and $\gcd(p,q,r) = 65^3 671^2$ fourth-power free.
Some somewhat simpler parametrizations are now known,
both for Euler's general problem and for this special case,
but I don't think I've seen one with an explanation that is
both elementary and well motivated.
A: 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:
$pqr=(6)^4$
$(p,q,r)=(3,6,72)$

A: 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,
$(p,q,r)=(30625,19600,400)$
