Find a,b,n so that (a^b)%n=x - c++

Question

Say I choose a value for x that can be between 0 and 2147483647. (Int32.MaxValue)
I am trying to figure out how I can find values for a,b,n so that (a^b)%n=x
I already know that I can use ModPow to verify the values, but I don't know how I can find a fitting a,b and n.
#include <iostream>
/// Calculate (a^b)%n
/// \param a The base
/// \param b The exponent
/// \param n The modulo
/// \return (a^b)%n
int ModPow(int a, int b, int n) {
long long x = 1, y = a;
while (b > 0) {
if (b % 2 == 1) {
x = (x * y) % n; // multiplying with base
}
y = (y * y) % n; // squaring the base
b /= 2;
}
return x % n;
}
int main() {
int x = 1337;
// How to find a,b,n so that (a^b)%n=x
int a = ?;
int b = ?;
int n = ?;
if(x == ModPow(a,b,n))
printf("ok");
return 0;
}

Answer 1

int n = 2147483647
int a = ModPow(x, 9241, n);
int b = 464773;
n = 231 − 1 is a prime number. So due to Fermat's little theorem, xn mod n = x and xn − 1 mod n = 1 (unless x = 0) so x2 n − 1 mod n = x, too. 2 n − 1 = 9241 × 464773. So (x9241 mod n)464773 mod n = x. Note that you need x < n for this to work; x = 2147483647 cannot work if n is a 31 bit (i.e. signed) integer, too.
It took me a while to get here; for a long time I've had this answer messing about with Carmichael numbers and the Carmichael function before I reached this easy solution. See edit history for details.

Answer 2

The modulus operator:
Yields the remainder given by the following expression, where e1 is the first operand and e2 is the second: e1 – (e1 / e2) * e2
Therefor whatever the max value of x is, n must be larger. Since you're validating with n as an int and you're specifying the range: 0 and numeric_limits<int>::max() that must be an exclusive range, and for n to be an int the only possible value it can take is: numeric_limits<int>::max().
With n forced our equation effectively becomes: ab = x.
We'll need to do a check here that x is not 1, if it is b = 0 and a can be anything in our legal range so we can arbitrarily pick a = 2. But baring this:
Our requirements are:
1 < a < x and a is an int
1 < b < x and b is an int
Given x, we can search for a combination of a and b that will fit as follows:
auto a = 0.0;
auto b = 1;
if(x == 1) {
a = 2.0;
b = 0;
} else {
while((a = pow(x, 1.0 / ++b)) > 2.0) {
double dummy;
if(modf(a, &dummy) == 0.0) {
break;
}
}
}
At this point, if a >= 2.0 then there is a valid solution to the problem. Now as you as you are probably well aware, pow is a very expensive function so this will likely take a very long time to execute for larger values of x, I'd personally suggest finding an a and b for every number for which such a pair exists and storing them in a map and doing a lookup on that.
Anyway this is a demonstration of working code:
Live Example