Think about the formula for a geometric progression:
which gives you two divisors straight off.
is an integral polynomial.
So, and are factors. If either of these are composite, then we have 3 prime factors readily. If they are both prime, then we observe , so we have another factor, .
It does not matter whether any of these factors are composite. If they are composite, they have at least one prime factor. Hence, we have at least 3 prime factors.
Thank you both for your answers!
Matt: could you tell me more on hwo your method involves the fact that the divisors have to be different? Eg. for b=3, we would have 8=2*2*2 in the denominator, so only one divisor, and how do we know how many there are on the left side?
Elemental: by integral polynomial you mean a polynomial with integer coefficients? If so, could you explain how you factor the f(a,b) or f(b,a) out? I don't really see it.
Sorry if I'm just being ignorant!
. For example, for , we have:
. In general, we have:
Notice the term is always there.
Since you said your primes are odd (since they aren't 2), then your expression:
is of the form . So basically I have factored it in two ways, yielding two distinct factors:
and . The polynomial I was referring to was simply:
I did not spell it out before because it wasn't important.
All you needed to know is that it WAS factorable.