Think about the formula for a geometric progression:
which gives you two divisors straight off.
Let's factor.
AND
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, .
So, .
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!
Yes, it has integer coefficients. This is a general polynomial. If we have an odd integer, , then we can ALWAYS factor an expression of the form:
. 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.
Make sense?
The above comments show that has at least three (prime) factors. But the question is asking for something much stronger than that, namely that is divisible by three different primes. For a proof of this difficult result, see melese's comment #3 in this thread.