# Modulo and prime numbers

by Texan
Tags: modulo, numbers, prime
 P: 3 $$f(x)$$ will give us the smallest integer $$m$$ such that $$y^m \equiv 1 \bmod{x}$$ given that x and y are coprime how would one go about showing that this function is multiplicative? I'm trying to do some Number Theory self study, and the problems I'm encountering are quite difficult to figure out from text alone. I'm guessing that the Chinese remainder theorem is applicable here. Also if we let p be prime then I know that f(p) will give the result of (p-1). This is basically proved with Euler's theorem. Is this true or is Euler's theorem not so easily applicable here? I guess the fact that here m is the smallest integer, where's in Euler's it's not the smallest, but it's easy to prove that this doesn't matter using groups.
 HW Helper Sci Advisor Thanks P: 7,939 How does y come in? Is it another given? If so, isn't it f(x, y)?
P: 891
 Quote by haruspex How does y come in? Is it another given? If so, isn't it f(x, y)?
I don't think so because y is dependent on x just as m is. I.E. y is any number coprime with x.

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## Modulo and prime numbers

 Quote by ramsey2879 I don't think so because y is dependent on x just as m is. I.E. y is any number coprime with x.
Rereading it, I think we're both wrong. It should say "for all y coprime to x". (Otherwise the later part about f(p) = p-1 is wrong.) In short, this f(x) is Euler's totient function.
 P: 39 i don't think it's Euler's totient function. for example, consider x = 37; y = 3. both prime, but the smallest m value is not 36, its 18.
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