
Is there any way to guess the answer without doing elaborating calculations?
Corrected after a comment of André Nicolas:
We have $4^5=1024>1016$. This shows that $y$ can only be $1$, $2$ or $3$. Now if $y=3$, then since $3^2\times 3^5>1016$, the only possible values of $x$ are $1$ and $2$. Similarly, if $y=2$, the only possible values of $x$ are $1,2,3,4,5$. Computing the products $xy$ and comparing them with the values proposed in the question, out of these seven options there remain only two: $(x,y)=(2,2)$ and $(x,y)=(5,2)$. Checking for $z^3$, out of these two possibilities there remains only one: $(x,y,z)=(5,2,6)$.
Now if $y=1$, then $xy=x$, and all we have to check is: which of the six values among $1016-(xy)^2$ are cubes. This gives one more triple $(x,y,z)=(4,1,10)$.
So the possible values of $xy$ (among indicated) are $4$ and $10$.
The correct solutions are 4 and 10. Denoting $f(x,y,z) = x^2y^5 + z^3$,
x y z f(x,y,z) xy
5 2 6 1016 10
4 1 10 1016 4
Here is the PERL code used to get this:
#!/usr/bin/perl -w
use strict;
my $z = 1;
my ($y5, $z3);
while (($z3 = $z**3) < 1016) {
my $x2y5 = 1016 - $z3;
my $y = 1;
while (($y5 = $y ** 5) <= $x2y5) {
next if $x2y5 % $y5;
my $x2 = $x2y5 / $y5;
if (int(sqrt($x2)) ** 2 == $x2) {
my $x = int(sqrt($x2));
print "($x, $y, $z) -> ", ($x**2) * ($y**5) + ($z**3), "; ", $x*$y, "\n";
}
} continue {
$y++;
}
} continue {
$z++;
}