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How to prove following statement :

For prime numbers $p$ greater than $3$, it is true that:

if $p=2^n-a$ and $a\equiv 1 \pmod 6$ then $p\equiv 1\pmod 3$

if $p=2^n+a$ and $a\equiv 5 \pmod 6$ then $p\equiv 1\pmod 3$

if $p=2^n-a$ and $a\equiv 5 \pmod 6$ then $p\equiv 2\pmod 3$

if $p=2^n+a$ and $a\equiv 1 \pmod 6$ then $p\equiv 2\pmod 3$

Maybe some conclusions from my previous question could be useful...

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1 Answer 1

up vote 1 down vote accepted

Observing the fact that $2^n =2,4 \mod 6$ and that $p \neq 0,3 \mod 6$ solves them..

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following your hint I have proved statement...thanks –  pedja Nov 17 '11 at 5:42

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