求解模23下的方程$x^{17}=17$，寻求通用解法及相关文献

Great question! You already found the unique solution (x=10) for (x^{17} \equiv 17 \mod 23) via brute force—let's break down the systematic way to solve equations of this form, starting with prime (q) (your case) and then touching on extensions.

Key Background: Multiplicative Groups Modulo Primes #

First, remember that for an odd prime (q), the non-zero elements modulo (q) form a cyclic multiplicative group of order (q-1). That means there exists some primitive root (g) such that every non-zero (a \mod q) can be written as (g^k) for some integer (k). This is the foundation of solving power congruences.

Step-by-Step Method for Prime (q) #

Let's walk through the process using your example ((q=23), (p=17), (a=17)) to make it concrete:

1. Verify if a solution exists #

A solution exists if and only if:
[a^{\frac{q-1}{\gcd(p, q-1)}} \equiv 1 \mod q]
For your case: (\gcd(17,22)=1), so we check (17^{22} \equiv 1 \mod 23)—which holds by Fermat's Little Theorem. Since (\gcd(p,q-1)=1), there's exactly one solution (matches your brute-force result).

2. Find a primitive root modulo (q) #

We need a generator (g) of the multiplicative group. For (q=23), (g=5) works (verify: (5^{22} \equiv1 \mod23), and no smaller exponent (<22) gives 1).

3. Express (a) as a power of the primitive root #

Find (k) such that (g^k \equiv a \mod q). For (g=5) and (a=17):
[5^7 = 78125 \equiv 17 \mod23]
So (k=7).

4. Convert to a linear congruence equation #

Let (x = g^m \mod q). Substitute into the original equation:
[g^{m \cdot p} \equiv g^k \mod q]
Since (g) has order (q-1), this simplifies to:
[m \cdot p \equiv k \mod (q-1)]
In your case: (17m \equiv7 \mod22).

5. Solve the linear congruence #

Since (\gcd(17,22)=1), 17 has an inverse modulo 22. We find (17 \times13 =221 \equiv1 \mod22), so the inverse is 13. Multiply both sides by 13:
[m \equiv7 \times13 \equiv91 \equiv3 \mod22]
Then (x =g^m=53=125 \equiv10 \mod23)—which is your solution!

Extensions to Non-Prime (q) #

If (q) is a prime power (e.g., (q=p^k)) or composite:

• For odd prime powers: The multiplicative group is still cyclic, so the same method applies (just use the primitive root modulo (p^k)).
• For (q=2^k) (k≥3): The multiplicative group isn't cyclic—it's isomorphic to (C_2 \times C_{2^{k-2}}). You'll need to solve the equation in each component of the group.
• For composite (q) with multiple prime factors: Use the Chinese Remainder Theorem—solve the equation modulo each prime power factor, then combine solutions.

Recommended References #

• Elementary Number Theory by David Burton: A great intro to primitive roots, power residues, and congruence equations with clear examples.
• A Course in Number Theory by Henry Davenport: A classic textbook that dives deeper into the algebraic structure behind these equations.
• Number Theory for Computing by Song Y. Yan: Focuses on practical algorithms, with code-friendly explanations for implementing these methods.
• Handbook of Applied Cryptography: Includes detailed sections on power residue computations, perfect for anyone working on cryptography-related applications.

内容的提问来源于stack exchange，提问作者Lau Mark