- Given an odd prime p, let's call the positive integer "p-tight" if all its prime factors are 1 (mod 2p).Theorem:if p does not divide a-1, then (a^p-1)/(a-1) is p-tight.(trivial proof follows from FLT)Conjecture:if p divides a-1, then (a^p-1)/(a-1)/p is p-tight.Any suggestions for a proof?Thanks a lot,Andrey
- Here is the extra step.

Let a = 1 + r*p. Then

S := (a^p-1)/(a-1)

= sum(k=0,p-1,a^k)

= sum(k=0,p-1,1+k*r*p) mod p^2

= p mod p^2,

since sum(k=0,p-1,k)=p*(p-1)/2 and p is odd.

Hence S/p = 1 mod p.

Happy now?That was so simple.Thanks.Andrey :-)