Mathematical Reflections 2014, Issue 1 - Problem J293

Problem:
Find all positive integers $x, y, z$ such that $$(x+y^2+z^2)^2-8xyz=1.$$

Proposed by Aaron Doman.

Solution:
We rewrite the equation as $$x^2+2x(y^2+z^2-4yz)+(y^2+z^2)^2-1=0.$$ Since $x$ must be a positive integer, the discriminant of this quadratic equation in $x$ must be non-negative, i.e. $$(y^2+z^2-4yz)^2-(y^2+z^2)^2+1 \geq 0,$$ which is equivalent to $$-8yz(y-z)^2+1 \geq 0,$$ which gives $yz(y-z)^2 \leq 1/8$. Since $y$ and $z$ are positive integers, it follows that $$yz(y-z)^2=0,$$ so $y-z=0$, i.e. $y=z$. The given equation becomes $(x-2y^2)^2=1$, which yields $x=2y^2 \pm 1$. Therefore, all the positive integer solutions to the given equation are $$(2n^2-1,n,n), (2n^2+1,n,n), \qquad n \in \mathbb{Z}^+.$$