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Given the Diophantine equation $$ x^2(8x-3)=y^2z, $$ is there a way to efficiently count the number of solutions that satisfy $x+y+z\leq n$, where $n$ is a fixed given integer?

Also, for any fixed $x$, is it possible to count all such solutions $(x,y,z)$ without having to explicitly find all the divisors of $x^2(8x-3)$?

A hint or a reference (if this is, in fact, easy) would be quite helpful.

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The estimate of this number from below can be done. Putting $x=k,\, y=k,\, z=8k-3, k \in \mathbb{N}$, we come to the inequality $k+k+8k-3 \le n$ having $[n/10]+O(1),\,\, n \to \infty,$ solutions. –  user64494 Aug 1 '13 at 6:03
An upper bound of this number can be obtained as the area of the part of the surface $x^2(8x-3)=y^2z$ in the first octant s. t. $x+y+z\leq n.$ –  user64494 Aug 1 '13 at 7:17
Reask this in MathOverflow. –  user64494 Aug 1 '13 at 7:25

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