Theory We know what Fermat's little theorem states. If $p$ is a prime number, then for any integer $a$, the number $a^p − a$ is an integer multiple of $p$. In the notation of modular arithmetic, this is expressed as \[a^{p}\equiv a{\pmod {p}}.\] So, essentially, for every $(a,m)=1$, ${a}^{\phi (m)}\equiv 1 \pmod {m}$. But $\phi (m)$ isn't necessarily the smallest exponent. For example, we know $4^{12}\equiv 1\mod 13$ but so is $4^6$. So, we care about the "smallest" exponent $d$ such that $a^d\equiv 1\mod m$ given $(a,m)=1$. Orders Given a prime $p$, the order of an integer $a$ modulo $p$, $p\nmid a$, is the smallest positive integer $d$, such that $a^d \equiv 1 \pmod p$. This is denoted $\text{ord}_p(a) = d$. If $p$ is a primes and $p\nmid a$, let $d$ be order of $a$ mod $p$. Then $a^n\equiv 1\pmod p\implies d|n$. Let $n=pd+r, r\ll d$. Which implies $a^r\equiv 1\pmod p.$ But $d$ is the smallest natural number. So $r=0$. So $d|n$. Show that $n$ divid...
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