# 2008 iTest Problems/Problem 91

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## Problem

Find the sum of all positive integers $n$ such that $x^3+y^3+z^3=nx^2y^2z^2$ is satisfied by at least one ordered triplet of positive integers $(x,y,z)$.

## Solution (credit to official solution)

WLOG, let $x \ge y \ge z$. We know that $x^3 + x^3 + x^3 \ge x^3 + y^3 + z^3$, so \begin{align*} 3x^3 &\ge nx^2 y^2 z^2 \\ x &\ge n \cdot \frac{y^2 z^2}{3} \end{align*} We also know that $x^3 + y^3 + z^3 \equiv 0 \pmod{x^2}$, so $y^3 + z^3 \equiv 0 \pmod{x^2}$. Since $x,y,z$ are positive, $y^3 + z^3 \ge x^2$. Thus, \begin{align*} 2y^3 &\ge y^3 + z^3 \ge x^2 \ge n^2 \cdot \frac{y^4 z^4}{9} \\ 2 &\ge n^2 \cdot \frac{y z^4}{9} \\ 18 &\ge n^2yz^4 \end{align*} Note that if $z > 1$, then $n^2yz^4 \ge 32$, so $z = 1$. Thus, $y \le \frac{18}{n^2}$, making $y \le 4$. Now perform casework on the values of $y$.

• If $y = 1$, then $2 \equiv 0 \pmod{x^2}$. Thus, $2$ must be a multiple of $x^2$. The only value of $x$ that works is $x = 1$, so $n = \tfrac{1+1+1}{1} = 3$.
• If $y = 2$, then $9 \equiv 0 \pmod{x^2}$. Thus, $9$ must be a multiple of $x^2$. The only value of $x$ greater than or equal to $2$ that works is $x = 3$, so $n = \tfrac{1+2+3}{6} = 1$.
• If $y = 3$, then $28 \equiv 0 \pmod{x^2}$. Thus, $28$ must be a multiple of $x^2$. However, there are no values of $x$ greater than or equal to $3$ that work.
• If $y = 4$, then $65 \equiv 0 \pmod{x^2}$. Thus, $65$ must be a multiple of $x^2$. However, there are no values of $x$ greater than or equal to $4$ that work.

The only values of $n$ that results in the equation solvable by positive integers is $n = 1$ and $n = 3$, so the answer is $1 + 3 = \boxed{4}$.