2023 USAJMO Problems/Problem 1

Problem

Find all triples of positive integers $(x,y,z)$ that satisfy the equation

$\begin{align*} 2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023 \end{align*}$

Solution 1

We claim that the only solutions are $(2,3,3)$ and its permutations.

Factoring the above squares and canceling the terms gives you:

$8(xyz)^2 + 2(x^2 +y^2 + z^2) = 4((xy)^2 + (yz)^2 + (zx)^2) + 2024$

Jumping on the coefficients in front of the $x^2$ , $y^2$ , $z^2$ terms, we factor into:

$(2x^2 - 1)(2y^2 - 1)(2z^2 - 1) = 2023$

Realizing that the only factors of 2023 that could be expressed as $(2x^2 - 1)$ are $1$ , $7$ , and $17$ , we simply find that the only solutions are $(2,3,3)$ by inspection.

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2023 USAJMO Problems/Problem 1

Problem

Solution 1