Question

Show that if$a^m+1$is composite if $a$and $m$are integers greater than 1 and $m$is odd. [ Hint: Show that $x+1$is a factor of the polynomial $x^m+1$if $m$is odd.]

Short answer

$a^m+1$ is composite

Step-by-step solution

Step 1

An integer$p$ larger than is called prime if the only positive factors of $p$ are 1 and $p$.

An integer that is not prime is called composite.

Step 2

Given: $a$ and $m$are integers greater than 1 and $m$is odd

To proof: $a^m+1$ is composite

PROOF

We add terms to $a^m+1$whose total is 0 (and thus does not affect $a^m+1$overall), then we regroup the terms using the distributive property.

$$\begin{gathered} a^m+1=a^m+\left(a^{m-1}-a^{m-1}\right)+\left(a^{m-2}-a^{m-2}\right)+\dots .+\left(a^2-a^2\right)+\left(a^1-a^1\right)+1 \\ =\left(a^m+a^{m-1}\right)-\left(a^{m-1}+a^{m-2}\right)+\left(a^{m-2}+a^{m-3}\right)+\dots -\left(a^2+a^1\right)+\left(a^1+1\right) \\ =a^{m-1}(a+1)-a^{m-2}(a+1)+a^{m-3}(a+1)+\dots -a(a+1)+(a+1) \\ =(a+1)\left(a^{m-1}-a^{m-2}+a^{m-3}+\dots -a+1\right) \end{gathered}$$

We then note that $\left(a+1\right)$is a factor of $a^m+1$ , while $\left(a+1\right)$is not 1 nor $a^m+1$, thus $a^m+1$is not prime and thus $a^m+1$is composite.

Note: If $m$is odd, then you would obtain $\left(-a^1+1\right)$ instead of $\left(a^1+1\right)$ in the second step, which is not divisible by $\left(a+1\right)$.