I need to find all primes of the form: $n^{5}-1$.
My attempt: First I have denoted $q=n^5-1$. It appears that $n$ is even, otherwise, if $n$ is odd then $n^5$ is odd too, which means that $n^5-1$ is an even number bigger than 2. Therefore, suppose that $n=2k,\mathbb{N}\ni k >0$. Thus, we have that: $n^5-1=(2k)^5-1=32k^5-1$. If $k=1$, then we have $32-1=31$ and we get a prime number. Now I know that I can proceed and get all 32's multiples, but I don't see how to negate that there are no more prime of the given form. Thanks!