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...