Question regarding the prime factors of $2^{35} - 1$
I just wanted to make a few things clear;
1) It is true to state that this cannot be a Mersenne prime (A number of the form $2^r - 1$ where if this is prime then $r$ must be prime, for $r \in \mathbb{Z}$) as r is not prime, as the converse of the definition of a Mersenne prime is not necessarily true.
2) I went about solving it by assuming $\exists$ a prime factor of the form $70x + 1$ this would therefore be for the first option $x = 1$ and hence $71$
(I now know that this is only necessary when trying to prove that a value of $2^r - 1$ is not prime for $r$ prime. - as every prime factor of $2^r - 1$ is of the form $2xr + 1$, and in this case $r = 35$ which is not prime)
Basically I want to show that $2^{35} \equiv 1 $ mod $71$ (Which I showed to be true).
So 71 is a prime factor. I also noticed that $2^5 - 1$ and $2^7 - 1$ are also prime factors equal to 31 and 127 respectively (Does this have something to do with the fact that 5 and 7 are prime factors of 35 - if so what is this fact, or is/should it be trivial).
And the last prime factor I figured out from wolfram - alpha to be equal to $122921$.
I was wondering if there was actually a method to figure out all these prime factors without rigorous number crunching, as I am yet to learn of one yet.