May take exponent of both sides of a modular congruence? For instance, may I write
$$n^2\equiv-1\mod p \quad \rm as \quad(n^2)^{2k+1}\equiv (-1)^{2k+1}\mod p ?$$
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Mhenni Benghorbal
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Slavica
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If $p|(a-b)$ does $p|a^k-b^k$?
LASV
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p|a^k-b^k as a^k-b^k can be written as (a-b)times something? – Slavica Nov 04 '13 at 23:56
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If the answer to my question is yes, can you conclude that the result is true? Remember what it means for $a\cong b \mod p$. It means that $p|a-b$. – LASV Nov 04 '13 at 23:58
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The result is true. – Slavica Nov 04 '13 at 23:59
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Yup, it is true. There are more than one way to show it. The way I told you is one of them. – LASV Nov 05 '13 at 00:00
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Since I can always factor (n^2)^2k+1-(-1)^2k+1,right? – Slavica Nov 05 '13 at 00:01
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Sure, if $a\equiv b \mod n$ then for every polynomial $f\in {\mathbb{Z}}[x]$ you have $f(a)\equiv f(b) \mod n .$
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