Can the substitution be different in this proof of a Fermat theorem?

Question

I was reading one of the Fermat theorem proof from the wikipedia
Specifically the theorem: " If $2^{k}+1$ is an odd prime, then $k$ is a power of 2."
In the proof it says:

Substituting $a=2^{r}$,$b=-1$,$b=-1$, $m = s$ and using that $s$ is odd

I originally thought that we could have used $a=2^s$ and $m=r$ and it seemed to me it should work too, but I was wondering if I am misunderstanding something as that statement there using that s is odd I am not sure if it means something more and would create a problem with a different substitution

Answer 1

If $k\in \Bbb Z^+$ and if $k$ is not a power of $2$ then $k\ge 3$ and $k$ has an odd divisor $p>1.$ So $k=k'p$ with $p$ odd and $p\ge 3$ and $k'\in \Bbb Z^+.$ Then, because $p$ is odd, we have $$2^k+1=2^{pk'}+1=(2^{k'})^p+1\equiv (-1)^p+1\equiv (-1)+1\equiv 0 \pmod {2^{k'}+1}$$ so $2^{k'}+1$ is a divisor of $2^k+1$ with $1<2^{k'}+1<2^{k'\cdot 3}+1\le 2^{k'p}+1=2^k+1$

so $2^k+1$ is not prime.

We can also see that $2^{k'}+1$ is a proper divisor of $2^k+1$ from the algebraic identity $x^p+1=(x+1)(x^{p-1}-x^{p-2}+-...+1)$ when $p$ is odd and $p\ge 3$. E.g

$x^3+1=(x+1)(x^2-x+1).$

$ x^5+1=(x+1)(x^4-x^3+x^2-x+1).$

$x^7+1=(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)$, et cetera.

For example with $x=2^{k'}$ and if $p=5.$ Then $x>1$ so $$x^4-x^3+x^2-x+1=(x^4-x^3)+(x^2-x)+1>1.$$

Answer 2

Using your substitution of $a = 2^{s}$ and $m = r$ instead, then if $r = 1$, this means $s = k$ so this gives

$$a^{m} - b^{m} = \left(2^{k}\right)^{1} - (-1)^{1} = 2^{k} + 1 \tag{1}\label{eq1A}$$

This then just states $2^{k} + 1 \mid 2^{k} + 1$, so it doesn't help with checking whether or not $2^{k} + 1$ is prime. However, if $r \gt 1$, then if $r$ is odd, it will also work just like it does with $s$. If $r$ is even instead, though, this then gives

$$a^{m} - b^{m} = \left(2^{s}\right)^{r} - (-1)^{r} = 2^{rs} - 1 = 2^{k} - 1 \tag{2}\label{eq2A}$$

However, the value being checked on is $2^{k} + 1$ instead.

Since using an odd factor $\gt 1$ is required to prove the divisibility requirement of $2^{k} + 1$ by a smaller value than itself, and $s$ meets this requirement, it's simpler & easier to just use this rather than unnecessarily breaking the proof down into additional cases trying to use $r$ instead in some situations. This is most likely why the proof in Wikipedia is done as it's shown.