I want to prove following statement :
For prime numbers $p$ greater than $3$, it is true that:
$a)$ if $p=2^n-a$ and $a=6k+1$, then $n$ is an odd number.
$b)$ if $p=2^n+a$ and $a=6k-1$, then $n$ is an odd number.
$c)$ if $p=2^n-a$ and $a=6k-1$, then $n$ is an even number.
$d)$ if $p=2^n+a$ and $a=6k+1$, then $n$ is an even number.
where $n \in \mathbf{Z}^+, k\in \mathbf{Z}^\ast$.
Proof :
$a)$ Lemma $1$ : $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+....+b^{n-1})$
$ p=2^n-(6k+1)=2(2^{n-1}+1)-6k-3=2(2^{n-1}+1)-3(2k+1)$
Let's suppose that $n$ is even then $n-1$ is odd and by the Lemma $1$:
$ (2+1) \mid (2^{n-1}+1) $ and since $(2+1) \mid 3(2k+1) $ we may conclude that $p$ is composite number.
So, we have contradiction, therefore $n$ must be odd number.
$b)$ Similarly as case $a)$
$c)$ $p=2^n-(6k-1)=2(2^{n-1}-1)-6k+3=2(2^{n-1}-1)-3(2k-1)$
let's suppose that $n$ is odd then $n-1$ is even ,therefore
$2^{n-1}-1=2^{2t}-1=(2^t-1)(2^t+1)$ , so :
$3 \mid (2^{n-1}-1)$ and since $ 3 \mid 3(2k-1)$ we may conclude that $p$ is composite number.
So,we have contradiction,therefore $n$ must be even number.
$d)$ Similarly as case $c)$
Is this an acceptable proof?