Question

For which positive integers nis${\mathit{n}}^{\mathbf{4}}\mathbf{+}{\mathit{n}}^{\mathbf{4}}$ prime?

Short answer

n = 1

Step-by-step solution

Step 1

DEFINITIONS

adivides b if there exists an integer c such that b = ac

Notation: a | b

An integer p is called prime ifp > 1 and if the only positive factors of p are 1 and p

n is even if and only if n is divisible by 2 if and only if there exists an integer such that n = 2k

nis odd if and only if there exists an integer k such that n = 2k+ 1

Step 2

SOLUTION

We need to derive when${\mathit{n}}^{\mathbf{4}}\mathbf{+}{\mathbf{4}}^{\mathbf{n}}$ is prime, where is a positive integer.

When n = 1 then${\mathit{n}}^{\mathbf{4}}\mathbf{+}{\mathbf{4}}^{\mathbf{n}}\mathbf{=}{\mathbf{1}}^{\mathbf{4}}\mathbf{+}{\mathbf{4}}^{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{4}\mathbf{=}\mathbf{5}$ is prime.

When n is an even positive integer, then there exists a positive integerk such that n = 2k

$\begin{array}{r}{\mathbf{n}}^{\mathbf{4}}\mathbf{+}{\mathbf{4}}^{\mathbf{n}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{k}{\mathbf{)}}^{\mathbf{4}}\mathbf{+}{\left(2^2\right)}^{\mathbf{n}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}{\mathbf{2}}^{\mathbf{4}}{\mathbf{k}}^{\mathbf{4}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}\mathbf{n}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\mathbf{2}\left(2^3{k}^4+2^{2n-1}\right)\end{array}$

Since k and n are positive integers, ${\mathbf{2}}^{\mathbf{3}}{\mathit{k}}^{\mathbf{4}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}\mathbf{n}\mathbf{-}\mathbf{1}}$is an integer as well and thus ${\mathit{n}}^{\mathbf{4}}\mathbf{+}{\mathbf{4}}^{\mathbf{n}}$is then divisible by 2.This then implies that${\mathit{n}}^{\mathbf{4}}\mathbf{+}{\mathbf{4}}^{\mathbf{n}}$ is not prime when is an even positive integer.

Step 3

When n is an odd positive integer, then there exists a positive integer k such that$\mathit{n}\mathbf{=}\mathbf{2}\mathit{k}\mathbf{+}\mathbf{1}$

role="math" localid="1668597978384" $\begin{array}{r}{\mathbf{n}}^{\mathbf{4}}\mathbf{+}{\mathbf{4}}^{\mathbf{n}}\mathbf{=}{\left(n^2\right)}^{\mathbf{2}}\mathbf{+}{\left(2^2\right)}^{\mathbf{n}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}{\left(n^2\right)}^{\mathbf{2}}\mathbf{+}{\left(2^n\right)}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}{\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{a}\mathbf{b}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}{\left(n^2+2^n\right)}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{\times }{\mathbf{n}}^{\mathbf{2}}\mathbf{\times }{\mathbf{2}}^{\mathbf{n}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}{\left(n^2+2^n\right)}^{\mathbf{2}}\mathbf{-}{\mathbf{n}}^{\mathbf{2}}\mathbf{\times }{\mathbf{2}}^{\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}{\left(n^2+2^n\right)}^{\mathbf{2}}\mathbf{-}{\mathbf{n}}^{\mathbf{2}}\mathbf{\times }{\mathbf{2}}^{\mathbf{2}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{1}}\mathbf{}\\ \mathbf{=}{\left(n^2+2^n\right)}^{\mathbf{2}}\mathbf{-}{\mathbf{n}}^{\mathbf{2}}\mathbf{\times }{\mathbf{2}}^{\mathbf{2}\mathbf{k}\mathbf{+}\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}{\left(n^2+2^n\right)}^{\mathbf{2}}\mathbf{-}{\mathbf{n}}^{\mathbf{2}}\mathbf{\times }{\mathbf{2}}^{\mathbf{2}\mathbf{(}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{)}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}{\left(n^2+2^n\right)}^{\mathbf{2}}\mathbf{-}{\left(n\times 2^{k+1}\right)}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ {\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\left(n^2+2^n-n\times 2^{k+1}\right)\left(n^2+2^n+n\times 2^{k+1}\right)\end{array}\mathbf{}$

Since k and n are positive integers, ${\mathit{n}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{n}}\mathbf{-}\mathit{n}\mathbf{\cdot }{\mathbf{2}}^{\mathbf{k}\mathbf{+}\mathbf{1}}$and ${\mathit{n}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{n}}\mathbf{-}\mathit{n}\mathbf{\cdot }{\mathbf{2}}^{\mathbf{k}\mathbf{+}\mathbf{1}}$ are both integers and both greater than 1. This then implies that ${\mathit{n}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{n}}\mathbf{-}\mathit{n}\mathbf{\cdot }{\mathbf{2}}^{\mathbf{k}\mathbf{+}\mathbf{1}}$is a factor of${\mathit{n}}^{\mathbf{4}}\mathbf{+}{\mathbf{4}}^{\mathbf{n}}$ different from$\mathbf{1}\mathbf{}\mathbf{and}\mathbf{}{\mathit{n}}^{\mathbf{4}}\mathbf{+}{\mathbf{4}}^{\mathbf{n}}\mathbf{,}$ which implies that ${\mathit{n}}^{\mathbf{4}}\mathbf{+}{\mathbf{4}}^{\mathbf{n}}$is not prime.

We can then conclude that ${\mathit{n}}^{\mathbf{4}}\mathbf{+}{\mathbf{4}}^{\mathbf{n}}$with a positive integer is only prime when$\mathit{n}\mathbf{=}\mathbf{1}$