Check the validity of "If $a^m\equiv 1\pmod b$ and $a^n\equiv 1 \pmod b$ then $m$ and $n$ are never co-prime."

Question

I recently say, some arguments, in a problem. The arguments, made me ponder upon it for hours. The question was:

Find number of triplets of positive integers satisfying

$2^a-5^b\cdot 7^c=1$

Given options are $0 , 1 , 2$ or infinite.

The solution goes like this in the thread,https://math.stackexchange.com/questions/1827071/number-theory-problem-finding-triplets?noredirect=1 :

Note that $ 2^a\equiv 1\pmod 5$ implies $a\equiv 0\pmod4$, so write $a=4m$. Now $$5^b7^c=2^a-1=(2^{2m}-1)(2^{2m}+1)$$ and the two factors are coprime. Take it further, $$5^b7^c=(2^{m}-1)(2^m+1)(2^{2m}+1)$$ and all three factors on the right are pairwise coprime. So at most one of them is divisible by $5$ and at most one is divisible by $7$, meaning that the smallest one must be $=1$, i.e., $m=1$. This gives us $$5^b7^c=(2^1-1)(2^1+1)(2^2+1)=15$$ which is absurd.

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Note that we used $b\ge 1$ to conclude $2^a\equiv 1\pmod 5$. If we were to allow $b=0$, we find additional solutions such as $a=1$, $b=c=0$ or $a=3$, $b=0$, $c=1$ and beyond that enter the realm of Catalan's conjecture.

But I am particularly, confused, about this statement "Note that $ 2^a\equiv 1\pmod 5$ implies $a\equiv 0\pmod4$, so write $a=4m$." ? Is this sort of, always true? For example, if $a^m\equiv c\pmod n$ and $a^n\equiv c\pmod n$ such that (WLOG) $m<n$ then it is not always true, that $m|n.$ So, I want to ask, under which circumstances can we make these sort implication? What are the criterias (if any) that must be satisfied, to go with this analogous reasonings?

Answer 1

What is true in general is this :

If $d$ is the smallest positive integer such that $a^d = 1 (mod\text{ }b)$, then for any positive integer $m$ such that $a^m = 1 (mod\text{ }b)$, $d | m$.

To see this, apply the division algorithm to obtain integers $q, r$ such that $m = dq + r$, where $0 \leq r < d$.

Then, $1 = a^m = (a^d)^q . a^r = a^r (mod \text{ } b)$, which contradicts the minimality of $d$, unless $r = 0$, so that $d | m$.

Algebraically speaking,

$a^k = 1 (mod\text{ }b)$ means that in the group $\mathbb{Z}^{\times}_b$, the element $[a]$ satisfies $[a]^k =1$. Thus, if $d$ is the smallest positive integer for which $a^d = 1 (mod \text{ }b)$ (that is the order of $[a]$ is $d$, then $d|k$.