Let S(A) be {1,2,3,...,(A-1)} and S'(A,B) be the set of elements X from S(A) that satisfy X^B mod A=1. Assuming that A is prime, find the sum of all the elements of S'(A,B) in terms of A and B.

(In reply to Hint2 by Praneeth)

Let d=gcd(B, A-1), and let z be a primitive d-th root of unity mod A.  Then the sum we are looking for is f(A,B) = sum(z^k, k=0...d-1).  Multiplying the sum by z only rearranges the terms mod A, so f(A,B)=0 mod A if z is not 1 mod A, which happens iff d>1.  Otherwise, if d=1, then the sum is just 1.

Now, if d is even, then the solutions to X^d=1 come in pairs (X, A-X), and so the sum is A*d/2.

If d is odd, it's more complicated.  If d=3, then the sum is < 1 + 2(A-1), and since it's a multiple of A, it must be A itself.  That is, f(A,B)=A if gcd(B,A-1)=3.  But even for d=5, I don't know the closed form.  The largest possible value is f(A,5)=3A, which happens only for A={101,421,641,...}.  f(A,5)=A for A={31, 151, 311, 541, 691, 821, 911, ...}, and f(A,5)=2A for

A={11, 41, 61, 71, 131, 181, 191, 211, 241, 251, 271, 281, 331, 401, 431, 461, 491, 521, 571, 601, 631, 661, 701, 751, 761, 811, 881, 941, 971, 991,...}

but I don't see the pattern.

Posted by Eigenray on 2007-08-02 08:04:00