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