How to prove indirectly that if $42^n - 1$ is prime then n is odd?

Solution 1:

Note that $$42^{2k}-1=(42^k)^2-1=(42^k-1)(42^k+1)$$ where $1\lt 42^k-1\lt 42^k+1$.

Solution 2:

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)

Solution 3:

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 $n$.