The smallest positive solution of the congruence $ax \equiv 0 \pmod n$ is called additive order of a modulo $n.$
To prove: if $p$ is a prime number and $a$ is any integer such that $p\nmid a$, then the additive order of $a$ modulo $p$ is equal to $p$.
Below find embedded questions for understanding better the following proof:
Applying the division algorithm we have $a=pq+r$ where $0<r<p$ (Note: $r>0$ because if $r=0$ then $p\mid a$ which is not the case. Clearly $pa \equiv 0$ (mod $p$) so solution of $ax \equiv 0$ (mod $p$) would be $x = 0 +pk$ where $k$ is an integer and the first positive solution is $p$.
The proof could stop here: $p$ being first positive solution then $p$ is the additive order of $a$ but the proof continues by stating that additive order of $a$ would be at most $p$ and at least 2 since $a$ is not identity. Questions: (1) Why do we need to show that additive order of $a$ is at most $p$ (in fact it is equal to $p$ as derived above) and at least $2$ the fact $a$ is not identity? (2) What is meant by identity? That $a=1$?
Next the proof shows, by contradiction, that additive order of $a$ is not only at most $p$ it is equal to $p$. First we assume $a < p$ and there are some integers $s$ such that $a<s<p$. Now $(p\mid a)s = (pq+r)s=pqs+rs$. Now the proof states that $p$ divides $pqr$ (that is OK) and it also divides $rs$ !?! Question: At this stage of the proof how can one state or show that $p$ divides $rs$? Next the proof invokes the fact that $p$ is prime, $p$ divides either $r$ or $s$ but this is impossible because $r$ and $s$ are less than $p$ and we have a contradiction $a$ is not less than $p$ and the proof concludes that additive order of $a$ modulo $p$ is $p$.
I fail to see why showing that $a$ is not less than $p$ leads to the conclusion that additive order of $a$ modulo $p$ is $p$. Also, why bring the integer $s$ that is $a<s<p$ in that proof by contradiction?
Any clarification and/or correction(s) of the above proof would be much appreciated.