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$$2^{2m+1}-1 \equiv 2(-1)^m-1 \in {1,2}\pmod{5}$$ – Jack D'Aurizio Feb 27 '17 at 19:46
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Quelle est votre question exactement? – Théophile Feb 27 '17 at 19:47
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i need a proof of the => – djamel math Feb 27 '17 at 20:38
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$!!\bmod 5!:,\ \left[1\equiv 2^{\large m}\right]^{\large 2}\Rightarrow, 1\equiv 4^{\large m}!\equiv (-1)^{\large m}!\equiv -1,\Rightarrow, 5\mid 2,\Rightarrow!\Leftarrow\ \ $ – Bill Dubuque Jan 03 '19 at 21:08
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Hint :
$5$ is a prime number.
So, if a positive integer $d$ divides simultaneously $5$ and $2^n-1$ and if $d\neq 1$, then ...
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