Exercise 15.5.9

Proof. Assume that $p^2-1=2^k$ for some integer $k\ge 0$. Then $(p+1)(p-1)=2^k$, therefore the unicity of the decomposition in prime factors shows that $p-1$ and $p+1$ are powers of $2$. There are integers $s,t\ge 0$ such that

where $0\le s<t$, and $s+t=k$.

Then $2=(p+1)-(p-1)=2^t-2^s$, thus

If $s=0$, then $p=2$ and $p^2-1=3$, which is not a power of $2$, so $s\ge 1$, and $t-s\ge 1$, so that $2^{s-1},2^{t-s}-1$ are positive integers, whose product is $1$. This shows that $2^{s-1}=1$, thus $s=1$, and $p=1+2^s=3$.

Conversely, if $p=3$, then $p^2-1=8=2^3$.

To conclude, $p^2-1$ is a power of $2$ if and only if $p=3$. □