Number Theory – Solving Diophantine System pqr=a^4, p+q+r=b^4

diophantine equationselementary-number-theorynumber theory

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 !

Best Answer

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.

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