# 2006 iTest Problems/Problem U4

The following problem is from the Ultimate Question of the 2006 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.

## Problem

As $n$ ranges over the integers, the expression $n^4 - 898n^2 + 1$ evaluates to just one prime number. Find this prime.

## Solution

Note that $898$ is two away from $900$. We can add and subtract $2n^2$ in the original expression to help factorization while maintaining equality. \begin{align*} n^4 - 898n^2 + 1 &= n^4 + 2n^2 + 1 - 2n^2 - 898n^2 \\ &= n^4 + 2n^2 + 1 - 900n^2 \\ &= (n^2 + 1)^2 - (30n)^2 \\ &= (n^2 + 30n + 1)(n^2 - 30n + 1) \end{align*} If none of the two factors are equal to $1$ or $-1$, then $n^4 - 898n^2 + 1$ is a composite number. If either one of the factors is set to equal $-1$, then $n$ would not be an integer. If either one of the factors is set to equal $1$, then $n = -30,0,30$. Plugging in $n = 0$ results in $1$, while plugging in $n = 30$ or $n = -30$ results in $1801$. From the description (or doing a prime check), the wanted prime number is $\boxed{1801}$.