# Difference between revisions of "2006 UNCO Math Contest II Problems/Problem 4"

## Problem

Determine all positive integers $n$ such that $n^2+3$ divides evenly (without remainder) into $n^4-3n^2+10$ ?

By polynomial division, $\frac{n^4 - 3n^2 + 10}{n^2 + 3} = n^2 - 6 + \frac{28}{n^2 + 3}$. By default, $n^2 - 6 \in \mathbb{N}$ when $n \in \mathbb{N}$, so we must find all positive integral values of $n$ for which $\frac{28}{n^2 + 3}$ is also an integer. Listing out the factors of $28$, we see that $n^2 + 3 \in \{1, 2, 4, 7, 14, 28\}$. Looking for positive integral values that satisfy these conditions yields $\boxed{n \in \{1, 2, 5\}}$, as desired. $\blacksquare$