Method to find solution for $a^x \equiv \mod n$

Question

The congruence $5^x \equiv 1 \mod 36$ has a solution because $5$ and $36$ are relatively prime, i.e. $5$ and $2^23^2$ have no common factors.

Is there a method to find $x$?

All I can see is that $5^3 \equiv 5 \mod 6$.

Answer 1

Hint

$$(6-1)^n\equiv(-1)^n+(-1)^{n-1}6n\pmod{36}$$

Check for odd$(2m+1)$ & even$(2m)$ values of $n$

Answer 2

You could always use the Chinese Remainder Theorem for congruences like this.

For $$ 5^x\equiv 1 \ \bmod 36 $$

you should solve

$$ \begin{cases} 5^x\equiv 1 \ \bmod 4\\ 5^x\equiv 1 \ \bmod 9 \end{cases} $$

This way you have to deal with smaller moduli.

Answer 3

In other words you want to find the multiplicative order of $5$ mod $36$. By Euler's theorem this divides $\varphi(36) = 12$. Next step: try $12/2 = 6$ and $12/3 = 4$.

Answer 4

Hint $\ a\equiv 1\pmod{np}\ \Rightarrow\, a^{\large p}\equiv 1\pmod{np^2}\ $ by $\,a^{\large p}\!-1 = (\overbrace{a-1}^{\large np\,k})(\!\overbrace{a^{p-1}+\cdots+a^2+a+1}^{\large\ \ \equiv\ 1+\cdots+1\ \equiv\ p\,\cdot 1\ \equiv\ 0\pmod{\!p}\!\!\!\!\!}\!\!)$

${\rm thus}\,\ 5^{\large 2}\!\equiv 1\pmod{\!4\cdot 3}\Rightarrow 5^{\large 6}\!\equiv 1\pmod{\!4\cdot 3^2}$

Remark $ $ To learn more about this general idea see LTE = Lifting The Exponent