Mersenne numbers: if equation x + y + axy = b has no soln's in positive integers, a =

[Diego_D]

True or False:
If the equation x + y + axy = b has no solution in positive integers, where ab + 2 = 2^a, and a - a prime, then 2^a -1 - prime.

Decision:
x + y + axy = b
ax + ay + ax ^. ay + 1 = ab + 1
(ax + 1) (ay + 1) = ab + 1
(ax + 1) (ay + 1) = ab + 2-1
(ax + 1) (ay + 1) = 2^a -1
All would be good if done with this.
Can I find an answer in the discussions? I have some doubts that the equation has no solution in positive integers, but 2^a -1 - composite number.

 

[Limax]

Suppose a = 2. Then \(\displaystyle (2x+1)(2y+1)=2^2-1=3\) and it is impossible for both x and y to be positive since one of the factors on the LHS must be 1. Also \(\displaystyle 2^2-1=3\) is prime. So the statement is true for a = 2.

Now let a be an odd prime. I will use this result: if p is a prime dividing \(\displaystyle 2^a-1\), then \(\displaystyle p\equiv1\pmod a\). It is not hard to prove it but I shall leave it for now.

Clearly the product of two primes dividing \(\displaystyle 2^a-1\) is also \(\displaystyle \equiv1\pmod a\). It follows that if p is not the only prime dividing \(\displaystyle 2^a-1\) or divides it more than once, we could choose x and y such that \(\displaystyle ax+1=p\) and \(\displaystyle ay+1\) is the product of the other prime factors, and then x and y would be positive solutions to the given equation. Hence p must be the only prime factor of \(\displaystyle 2^a-1\) dividing it just once, meaning \(\displaystyle 2^a-1=p\) is prime.