GrimReaper42's blog

By GrimReaper42, history, 4 weeks ago, In English,

I have found that the answer to my question is

a^((b^c) % prime — 1) % prime

But I don't know a proof, can anyone tell me the proof or give me a tutorial to read it. Thanks

 
 
 
 
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4 weeks ago, # |
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Fermat's little theorem states that for a prime $$$p$$$ and any integer $$$a$$$ such that $$$0 < a < p$$$,

$$$ a^{p-1} \equiv 1 \implies a^b \equiv a^{b \bmod {p-1}} \pmod{p} $$$

In other words, the powers of $$$a$$$ modulo $$$p$$$ repeat with a period of $$$p-1$$$.

This is further generalized by Euler's theorem which implies the powers of $$$a$$$ repeat with a period of $$$p-1$$$. This can be even further generalized with Lagrange's theorem for groups.

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    4 weeks ago, # ^ |
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    Yeah, Fermat's Little Theorem basically finishes off this problem, since we have a period p-1, now all we need to do is find $$$b^c \mod p-1$$$ (just use binary exponentiation) and then find $$$a^{(\text{this number})} \mod p$$$. One more generalized formula is Euler's Totient Theorem, which states that $$$a^{\phi(x)} \equiv 1$$$ $$$(\text{mod }x)$$$ where $$$\gcd(a, x)=1$$$.

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4 weeks ago, # |
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https://www.geeksforgeeks.org/find-abm-where-b-is-very-large/

You can use above link for better understanding.