Is the following statement true and how to prove it? \begin{align} (a^2)^{3N} \equiv a^2 \mod{p} \end{align}
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what is N? $N \epsilon N$ ? – Nir Agami Apr 15 '14 at 11:24
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1for a general $N$, $a$ and $p$ this is most definitely not true – Jack Yoon Apr 15 '14 at 11:25
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N & a are positive integers greater than 1. – SAMARA Apr 15 '14 at 11:30
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A counter example: I assume that N is a generic natural number and p is any prime number.
Let's say $N=2 , p=7$ $a=2$
$$(2^2)^{3*2}=4096$$ and $$4096(mod7)=1$$
and $1\neq4(mod7)$ a contradiction.
Nir Agami
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Since $(a^2)^{3N}\neq a^2$ if $a\neq 1$, it is obvious that the statement cannot hold for arbitrary values of $p$. For example, taking any $p>(a^2)^{3N}$ will provide a counterexample.
5xum
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