2006 UNCO Math Contest II Problems/Problem 4

Problem

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

Solution

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. $\blacksquare$

2006 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2006 UNCO Math Contest II Problems/Problem 4

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