Let A be an integer, P an odd prime and n=3 be the smallest integer for which A^n  1 is divisible by P.
Determine the smallest integer m for which (A+1)^m  1 is divisible by P.
(In reply to
Some more thoughts by Dej Mar)
obviously m can't be negative or else (A+1)^m would not be an integer.
Also m can not be 1 because we would then have that A is a multiple of P and if that was true we would have that A=Pk and from the first part we would get A^31=Pt or (Pk)^31=Pt or
t=[(Pk)^31]/P=P^2k^3(1/P) and thus t would not be an integer thus giving us a contradiction therefore m can not be 1
now where it gets tricky is if we allow m=0 because in that case the second part breaks down to saying that 0 is divisble by P and since technically 0 is divisible by any number that would hold true and so it would seem that 0 is the smallest integer m that is being looked for.
So in conlcusion if you allow m=0 then 0 is the answer otherwise m=2 is the answer as shown by Dej Mar

Posted by Daniel
on 20070109 11:02:29 