This is the second part to a question in the introductory proof book How to Solve It. In the first part I factored 2^15 - 1 = 32767 into 31(1057) and I found that using this xy = (2^b − 1) · (1 + 2^b + 2^2b + · · · + 2^(a−1)b ) , where xy = n and the n is from 2^n-1 which can possibly result in a prime number if n is prime, but will not equal a prime number if n is not prime.
the second question is this - Find an integer x such that 1 < x < 2^32767 − 1 and 2^32767 − 1 is divisible by x. I know that 32767 = 31(1057) and so either 31 or 1057 is greater than 1 and less than 2^32767-1, but I'm lost how to find the integer that divides into 2^32767-1 and would appreciate any help.
You use the same logic. You have identified that $2^a-1|2^{ab}-1$, so use your factorization of $32767=31\cdot 1057$ to say that $2^{32767}-1$ is divisible by $2^{31}-1$ or $2^{1057}-1$. In fact $1057=7\cdot 151$ so you have a couple other factors available as well.