Assume $\gcd(a, m) > 1$. Show that there cannot possibly exist a power $k$ such that $a^k\equiv 1 \bmod m$.

Question

i have no idea where to start to show this. i'm assuming that i should probably use Fermat little theorem and division algorithm. but i don't know where to start.

Answer 1

Hint $\ a^k\equiv 1\pmod m\,\Rightarrow\,jm + a^k = 1\,$ so $\,d\mid m,a\,\Rightarrow\, d\mid jm + a^k = 1,\,$ if $\ k> 0$

And $\ \ \, ak\equiv 1\pmod m\,\Rightarrow\,jm + ak = 1\,$ so $\,d\mid m,a\,\Rightarrow\, d\mid jm + ak = 1$

Note that the first is a special case of the second, since $\,a(a^{k-1})\equiv 1$

Generally, units (= invertibles = divisors of $1$) cannot be zero-divisors (except in a trivial ring).

Answer 2

Suppose there is $k>0$ such that $a^k\equiv 1\pmod{m}$. Then there is some integer $z$ such that $$ a^k=1+mz\implies [a^{k-1}]a+[-z]m=1\iff ra+sm=1\text{ where }r=a^{k-1}, s=-z. $$ This is a Bezout's identity for $a$ and $m$ that implies $\text{gcd}(a,m)=1$.

Answer 3

If such $k$ exists then the eqation $ax\equiv 1\bmod m$ has a solution (i.e., $x=a^{k-1}$). while the equation $ax\equiv 1\bmod m$ has a solution if and only if $\gcd(a,m)=1$.