# Difference between revisions of "1989 AIME Problems/Problem 9"

## Problem

One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that $133^5+110^5+84^5+27^5=n^{5}$. Find the value of $n$.

## Solution

Note that $n$ is even, since the $RHS$ consists of two odd and two even numbers. By Fermat's Little Theorem, we know ${n^{5}}$ is congruent to $n$ modulo 5. Hence,

$3 + 0 + 4 + 7 \equiv n\pmod{5}$
$4 \equiv n\pmod{5}$

Continuing, we examine the equation modulo 3,

$-1 + 1 + 0 + 0 \equiv n\pmod{3}$
$0 \equiv n\pmod{3}$

Thus, $n$ is divisible by three and leaves a remainder of four when divided by 5. It's obvious that $n>133$, so the only possibilities are $n = 144$ or $n \geq 174$. It quickly becomes apparent that 174 is much too large, so $n$ must be $144$.