如何求解模方程$x^y = a \mod{b}$？大素数模下$x^{37}=a \mod{b}$求解困境

Alright, let's work through this problem step by step—you were on the right track with Fermat's Little Theorem (FLT), but let's unpack how to apply it properly even for huge (b).

First, let's cover the quick edge case: if (a \equiv 0 \pmod{b}), then (x \equiv 0 \pmod{b}) is the only solution—done. For all other cases where (a) isn't divisible by (b), we lean into FLT and modular arithmetic to find (x).

Key Background from Fermat's Little Theorem #

Since (b) is prime, FLT tells us that for any (x) not divisible by (b):
(x^{b-1} \equiv 1 \pmod{b})
This means exponents in modular arithmetic modulo (b) can be reduced modulo (b-1) when working with powers of (x). Our core goal is to find an exponent (k) such that (37k \equiv 1 \pmod{b-1})—this (k) will let us compute (x) as (a^k \pmod{b}), because:
((x^{37})k = x^{37k} = x^{1 + m(b-1)} = x \cdot (x^{b-1})m \equiv x \cdot 1^m = x \pmod{b})

Step-by-Step Solution #

1. Compute the GCD of 37 and (b-1) #

Since 37 is a prime number, the gcd can only be 1 or 37:

• If (\gcd(37, b-1) = 1): 37 has a unique modular inverse modulo (b-1)
• If (\gcd(37, b-1) = 37): We need to check if a solution exists first, then find all possible solutions

Case 1: (\gcd(37, b-1) = 1) #

This is the straightforward scenario:

• Use the extended Euclidean algorithm to find (k) such that (37k \equiv 1 \pmod{b-1}). Even for massive (b) (think thousands of digits), this algorithm runs efficiently in (O(\log b)) time.
• In code (Python example), you can use the built-in pow function for this inverse: k = pow(37, -1, b-1) (works because 37 and (b-1) are coprime)
• Compute (x \equiv a^k \pmod{b}) using modular exponentiation: x = pow(a, k, b)
• This (x) is the unique solution modulo (b)

Case 2: (\gcd(37, b-1) = 37) #

Here, we first need to verify if a solution exists:

1. Check solvability: Compute (a^{(b-1)/37} \pmod{b}). If this is not congruent to 1 modulo (b), there are no solutions to the equation.
2. If solutions exist:
   • Let (s = (b-1)/37). Since (\gcd(37, s) = 1) (we already factored out all 37s from (b-1)), 37 has an inverse modulo (s)—let's call this (k_0) (find it via extended Euclidean or pow(37, -1, s)).
   • A single solution is (x_0 \equiv a^{k_0} \pmod{b})
   • To find all 37 solutions, multiply (x_0) by each 37th root of unity modulo (b). A 37th root of unity is any (\omega) such that (\omega^{37} \equiv 1 \pmod{b}). You can get this by taking a primitive root (g) modulo (b) (a generator of the multiplicative group) and setting (\omega = g^s) (since (\omega^{37} = g^{37s} = g^{b-1} \equiv 1 \pmod{b})).
   • All solutions are (x \equiv x_0 \cdot \omega^t \pmod{b}) for (t = 0, 1, ..., 36)

Why This Works Even for Large (b) #

All the operations here—gcd calculation, extended Euclidean algorithm, modular exponentiation—are efficient even for extremely large primes (like those used in cryptography). Modular exponentiation uses exponentiation by squaring, which runs in (O(\log k)) time regardless of how big (b) is.

内容的提问来源于stack exchange，提问作者Jackson Blankenship