Fix an integer k > 0, and consider the function Sigma k : N rightrrow N by sigma k Prove that sigma k (n) (mod p) for all primes p. Solution We have k(n) = (d1)k + (d2)k + ... + (dj)k where d1, d2, ..., dj are all the divisors of n. Using the same formula, pk(n) = (d1)pk + (d2)pk + ... + (dj)pk = (d1p)k + (d2p)k + ... + (djp)k. By Fermat\'s Little Theorem, for any prime p, ap = a (mod p). So each of the d1p, d2p, ..., djp is congruent to d1, d2, ..., dj (mod p) respectively, and therefore (d1p)k + (d2p)k + ... + (djp)k is congruent to (d1)k + (d2)k + ... + (dj)k (mod p). This shows that pk(n) = k(n) (mod p) (where \"=\" denotes congruence). The proof that pik(n) is congruent to k(n) (mod p) is a simple extension of this. We have pik(n) = (d1)p^i k + (d2)p^i k + ... + (dj)p^i k = (d1p^i)k + (d2p^i)k + ... + (djp^i)k = (d1p*p*p*...*p)k + (d2p*p*p*...*p)k + ... + (djp*p*p*...*p)k (where each exponent is p multiplied by itself i times) = ((d1p)p*p*...*p)k + ((d2p)p*p*...*p)k + ... + ((djp)p*p*...*p)k which by Fermat\'s Little Theorem, is congruent to ((d1)p*p*...*p)k + ((d2)p*p*...*p)k + ... + ((dj)p*p*...*p)k (mod p), with each exponent having one less p multiplied in, since each d^p reduced to d (mod p) by Fermat\'s Little Theorem. We can keep applying Fermat\'s Little Theorem to find that each pik(n) is congruent to pi-1k(n), and therefore all will eventually be congruent to pk(n) and therefore congruent to k(n)..

1. Fix an integer k > 0, and consider the function Sigma k : N rightrrow N by sigma k Prove that
sigma k (n) (mod p) for all primes p.
Solution
We have
k(n) = (d1)k + (d2)k + ... + (dj)k
where d1, d2, ..., dj are all the divisors of n. Using the same formula,
pk(n) = (d1)pk + (d2)pk + ... + (dj)pk
= (d1p)k + (d2p)k + ... + (djp)k.
By Fermat's Little Theorem, for any prime p,
ap = a (mod p).
So each of the d1p, d2p, ..., djp is congruent to d1, d2, ..., dj (mod p) respectively, and therefore
(d1p)k + (d2p)k + ... + (djp)k is congruent to (d1)k + (d2)k + ... + (dj)k (mod p). This shows that
pk(n) = k(n) (mod p)
(where "=" denotes congruence).
The proof that pik(n) is congruent to k(n) (mod p) is a simple extension of this. We have
pik(n) = (d1)p^i k + (d2)p^i k + ... + (dj)p^i k
= (d1p^i)k + (d2p^i)k + ... + (djp^i)k
= (d1p*p*p*...*p)k + (d2p*p*p*...*p)k + ... + (djp*p*p*...*p)k
(where each exponent is p multiplied by itself i times)
= ((d1p)p*p*...*p)k + ((d2p)p*p*...*p)k + ... + ((djp)p*p*...*p)k
which by Fermat's Little Theorem, is congruent to
((d1)p*p*...*p)k + ((d2)p*p*...*p)k + ... + ((dj)p*p*...*p)k (mod p),
with each exponent having one less p multiplied in, since each d^p reduced to d (mod p) by
Fermat's Little Theorem. We can keep applying Fermat's Little Theorem to find that each pik(n)
is congruent to pi-1k(n), and therefore all will eventually be congruent to pk(n) and therefore
congruent to k(n).