mathschallenge.net

Problem

Given that $n$ is a natural number, when is $n^4 + 4$ prime?

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Solution

$$\begin{aligned}n^4 + 4 & = n^4 + 4n^2 + 4 - 4n^2\\& = (n^2 + 2)^2 - (2n)^2\\& = \left( (n^2 + 2) + 2n \right)\left( (n^2 +2) - 2n \right)\\& = (n^2 + 2n + 2)(n^2 - 2n +2)\end{aligned}$$

As this represents a product it can only be prime if one of the factors is equal to 1.
Clearly $n^2 + 2n + 2 \gt 1$ for $n \ge 1$.

$$\begin{aligned}\therefore n^2 - 2n + 2 & = 1\\n^2 - 2n + 1 & = 0\\(n - 1)^2 & = 0 \Rightarrow n = 1\end{aligned}$$

When $n = 1$ we get the prime, $n^4 + 4 = 5$. Hence this is the only value of $n$ for which $n^4 + 4$ is prime.

Explain why $n^4 + 4$ will always divide by two distinct numbers of the form $k^2 + 1$ for $n \gt 1$. For examle, $5^4 + 4 = 629,$ which is divisible by $6^2 + 1 = 37$ and $4^2 + 1 = 17$.