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If $a$ and $P$ are two relatively prime positive integers, such that $a < P$, and $m$, $n$ are natural numbers less than $P$, will $(m*a)$ modulo $P$ ~= $(n*a)$ modulo $P$, always?

That is, will the multiples of $a$ modulo $P$, be not equal always? or will all the multiples of $a$ generate all the elements of the finite field GF(p). Note that P can be non-prime.

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in the first line please append, 'a<P', and 'm','n' are natural numbers less than P. – TGM Oct 1 2010 at 16:56
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 Do you want $n \neq m$? And should ~= be $\neq$? If both is the case, it is clearly correct, since $a$ is a unit in $\mathbb{Z}/P\mathbb{Z}$. Also, $GF(p)$ does not make sense if $P$ is not prime; you probably mean $\mathbb{Z}/P\mathbb{Z}$. (Is $p = P$, anyway?) In any case, I'm not sure if this question belongs to this site; maybe the math section on stackoverflow.com is better suited. – felix Oct 1 2010 at 17:15
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 Most things are true modulo arithmetic. – Cam McLeman Oct 1 2010 at 17:30
The question could do with being made more precise; but in any case it might belong better on math.stackexchange.com – Yemon Choi Oct 1 2010 at 18:37

closed as too localized by HW, Tony Huynh, Robin Chapman, Yemon Choi, Ryan Budney Oct 1 2010 at 18:58

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