Divisors of (X^a-1)/(X-1) x is an odd prime, a is odd integer

When factoring a number like (37^3-1)/36 = 3*7*67, I've noticed that almost all have at least one prime factor larger than x (e.g. 67 > 37). I would like to know for what values of x,a are ALL of the prime factors of (X^a-1)/(X-1) less than X. For example (79^3-1)/78 = 3*7^2*43 and 43 < 79 so one example is x = 79, a = 3. My math education level is first year of high school so a transparent explanation, if possible, would be great. I understand basic congruences.

Re: Divisors of (X^a-1)/(X-1) x is an odd prime, a is odd integer

Certainly. has the form , correct? So what you're really asking for is the set of integers of the form

which are such that primes , . This is not a high school level problem, particularly for U.S. students, but it is not beyond approach. Let me put it this way for you, which should lead you to the answer:

What you're trying to avoid is that there is a prime such that [ ], [ ], , [ ] where [ ]. So try looking at a few cases that solve the system of equations you're looking for them to solve and make a couple of hypotheses (although it should be moderately apparent, and if not, the OEIS is always a good tool, if it wouldn't feel too much like cheating. But don't only look at primes for values of or you will not find the sequence!). You will find that there is a certain class of numbers that admits the property which you seek, and a proof by contradiction follows without too much trouble. It is a much easier problem algebraically than when using strictly elementary tools, but I see how to go about it using only elementary techniques. Another hint: Ignore the '1' and try to avoid [ ].

The proof is not short, but I will come back in a week or two and write it out for you if you still have not come up with the solution.