TITLE: How to prove indirectly that if $42^n - 1$ is prime then n is odd?
QUESTION [10 upvotes]: I'm struggling to prove the following statement:

If $42^n - 1$ is prime, then $n$ must be odd.

I'm trying to prove this indirectly, via the equivalent contrapositive statement, i.e. that if $n$ is even, then $42^n - 1$ is not prime.
By definition, for every even number $n$ there exists an integer $k$ with $n = 2k$. We substitute and get
$$42^n - 1 = 42^{2k} - 1 = (42^2)^k - 1.$$
Now, how do I prove that $(42^2)^k - 1$ isn't a prime number? Is this even the right way to approach this proof?

REPLY [11 votes]: Note that 
$$42^{2k}-1=(42^k)^2-1=(42^k-1)(42^k+1)$$
where $1\lt 42^k-1\lt 42^k+1$.

REPLY [7 votes]: Note that $$\begin{align}42^n - 1 &\equiv 1 - 1 \\&\equiv 0 \pmod{41}\end{align}$$
so the only way for $42^n - 1$ to be a prime is for $n$ to be $1$.
In general, for $a^n - 1$ to be a prime, where $a, n \in\mathbb{Z}^+$, either $a = 2$ or $n = 1$.
(Not sure if this counts as indirect, but you could turn it into some form of contradiction)

REPLY [6 votes]: By the binomial theorem, $42^n = (43-1)^n=43a+(-1)^n$.
If $n$ is even, then $42^n-1$ is a multiple of $43$.
On the other hand, $42^n = (41+1)^n=41b+1$, and so $42^n-1$ is always a multiple of $41$. Thus, $42^n-1$ is not prime if $n>1$, regardless of the parity of $n$.