Use Fermat's Little Theorem to solve the equation $x^{13} \equiv 2\;(\mod\ 23)$

Question

I have to use Fermat's Little Theorem to solve the equation $$x^{13} \equiv 2\;(\mod\ 23)$$ But I have no idea how to do so. I understand that $2^{23} \equiv 2\;(\mod\ 23)$. But I don't know how to solve what the question is asking.

Answer 1

Fermat's little theorem says that if $x$ is relatively prime to $23$, then $x^{22}\equiv1\pmod{23}$.

Moreover, $$13\times17=221=22\times10+1.$$ So if $$x^{13}\equiv2\pmod{23},$$ then $$(x^{13})^{17}\equiv x^{22\times10+1}\equiv x^{1}\equiv2^{17}\pmod{23}.$$ Now $$\begin{align}2^2&\equiv4\\2^4&\equiv16\\2^8&\equiv256\equiv3\\ 2^{16}&\equiv9\end{align}.$$ So $2^{17}\equiv18\pmod{23}$. As a consequence, we find that $18$ is a solution to $x^{13}\equiv2\pmod{23}$.

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Hope this helps.

Answer 2

Another approach:

⇒ $x^n=2+23k$

This equation can have numerous solutions; for example (k=1, n=2, x=5) or ( k=13, n=2, x=18) which satisfies following congruence:

$x^2≡2 \mod 23$

If (x, 23)=1 then:

$x^{22}=(x^{11})^2 ≡ 1 \mod 23 $,⇒ $x^{11}≡ 1 \mod 23$

⇒ $x^{11}\times x^2=x^{13}≡2 \mod 23$

For example a set of solutions for congruence $a^n ≡ 2 \mod 23$is:

$5^2, 5^{13}, 5^{13+11}=5^{24}, 5^{35}, . . .$

Another set of solution is:

$18^2, 18^{13}, 18^{24}, 18^{35}, . . .$

so for your question x=5 and x=18 and generally all number which their squares have remainder 2 when divided by 23 are solutions.

Answer 3

If curious you may like using discrete log: $$x^{a} \equiv b\pmod{p} \Rightarrow a\log_rx\equiv \log_rb\pmod{p} \Rightarrow x\equiv r^{\frac{\log_rb}{a}}\pmod{p}$$

Sadly this works only when $p$ is prime.

Answer 4

Let us find $a,b$ such that $13a+22b=1,$

By Observation $b=3, a=-5$

Using Fermat's Little Theorem,

$$x=x^{22(3)-5(13)}=(x^{22})^3(x^{13})^{-5}\equiv1^3\cdot2^{-5}\pmod{23}$$

Let us find $c,d$ such that $-32c+23d=1\implies c\equiv32^{-1}\pmod{23}$

Use my method from Solving a Linear Congruence

Or using Fermat's Little Theorem, $2^{-5}\equiv2^{17}\pmod{23}$