Let $n$ a natural number non zero
Prove that: $$n^5 \equiv n\mod 30$$
I have tried to use fermat theorem and little fermat i can't find any result
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2$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=(n-1)n(n+1)(n^2+1)$. The first three factors guarantee a factor of 6. Consider what happens modulo 5. – symplectomorphic Apr 02 '17 at 23:57
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First, $n^5 - n$ is a multiple of 5 according to Fermat's Little Theorem. Second, $n^5 - n$ is a multiple of 2, because $n^5$ has the same pairty as $n$. Finally, $n^5 - n$ is a multiple of 3: if $n$ is divisible by 3 then $n^5 - n$ is also divisible by 3; if $n \equiv 1$ then $n^5 \equiv n \equiv 1$ and if $n \equiv 2$ then $n^5 \equiv n \equiv 2$.
Thus, $n^5 - n$ is a multiple of $5 \cdot 2 \cdot 3 = 30$.
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when you write "$n^5$ has the same oddity as $n$," the word you want is "parity," not "oddity." – symplectomorphic Apr 03 '17 at 00:04
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