Inverse of A*X MOD (2^N)-1

Question

Given a function y = f(A,X):

unsigned long F(unsigned long A, unsigned long x) {
    return  ((unsigned long long)A*X)%4294967295;
}

How would I find the inverse function x = g(A,y) such that x = g(A, f(A,x)) for all values of 'x'?

If f() isn't invertible for all values of 'x', what's the closest to an inverse?

(F is an obsolete PRNG, and I'm trying to understand how one inverts such a function).

• Updated
  If A is relatively prime to (2^N)-1, then g(A,Y) is just f(A-1, y).
  If A isn't relatively prime, then the range of y is constrained... Does g( ) still exist if restricted to that range?

Answer 1

You need to compute the inverse of A mod ((2^N) - 1), but you might not always have an inverse given your modulus. See this on Wolfram Alpha.

Note that

A = 12343 has an inverse (A^-1 = 876879007 mod 4294967295)

but 12345 does not have an inverse.

Thus, if A is relatively prime with (2^n)-1, then you can easily create an inverse function using the Extended Euclidean Algorithm where

g(A,y) = F(A^-1, y),

otherwise you're out of luck.

UPDATE: In response to your updated question, you still can't get a unique inverse in the restricted range. Even if you use CookieOfFortune's brute force solution, you'll have problems like

G(12345, F(12345, 4294967294)) == 286331152 != 4294967294.