Thread: Pigeonhole problem

Prove that there exists a positive integer n so that 44^n — 1 is divisible
by 7.

There's no real need for a pigenhole principle here (look for little Fermat's theorem), but let's do it your way.
Consider the eight numbers . There are only 7 possible remainders in the division of these numbers by 7 (namely 0, 1,..., 6), hence at least two of these numbers have the same remainder modulo 7: this results from a pigeonhole principle. Let's say it is and [tex]44^n[/Math], [tex]n<m[/Math]. Then, for some , and , hence . Because , this gives , so that divides . In addition, and are relatively prime, hence we deduce that divides . And . We are done.

thank you very much

Originally Posted by dajaka

SOLVED, THANX

It's not good form to delete questions. Now other members can't view it at a later date and perhaps learn something.

@Laurent: A good reason to always quote the question when replying .......

ok sorry
here is the problem again...