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The full question is this:

find an integer $0<n<23$ such that, if $x∈Z$ is a solution to

$103x^5 \equiv 1 \mod 23$,

then

$x \equiv n \mod 23$.

I'm not sure how to approach it, though I think Fermat's Little Theorem and/or Euclid's algorithm will help.

I thought maybe the fact that $103 (-2) \equiv 1 \mod 23$ could be helpful too, but I'm not sure.

J. W. Tanner
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3 Answers3

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$103x^5\equiv1\bmod23\iff 103^9x^{45}\equiv1\bmod23$

$\overset{Fermat}\iff 103^9x\equiv1\bmod23\iff x\equiv(-2)^9\bmod23.$

Can you take it from here?

J. W. Tanner
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  • I used the fact you gave that $103(-2)\equiv1\bmod23$ – J. W. Tanner May 11 '20 at 18:04
  • Presumably you also used $,1/5\equiv 9\pmod{22},$ as I explain in my answer (or did you use some other method?) – Bill Dubuque May 11 '20 at 20:41
  • @Gone: yes, I raised to the $9^{th}$ power because $5\times9\equiv1\bmod22$; I easily solved in my head $5y\equiv1\equiv23\equiv45\bmod22\iff y\equiv9\bmod22$ – J. W. Tanner May 11 '20 at 20:52
  • But you didn't explain why you chose $9$ in your answer - you simply pulled it out of a hat like magic. How is the OP supposed to know what to do for similar cases if you don't explain the concept behind it? This is usually the stumbling block in problems like this. – Bill Dubuque May 11 '20 at 21:29
  • I figured OP would see why raising to the power of $9$ worked so well to solve for $x$ and would learn from this example – J. W. Tanner May 11 '20 at 22:57
  • One can only hope. But given the frequent questions here showing confusion about plain modular inverses it shouldn't come as a surprise that it may be even more confusing when happening in exponents (with a different modulus / period). Better to be explicit. – Bill Dubuque May 11 '20 at 23:03
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Here is a variation of J. W. Tanner's answer, probably more suitable for hand computation.

$103x^5\equiv1\bmod23$

$\iff 11x^5\equiv1\bmod23$

$\iff x^5 \equiv -2 \bmod23$

$\iff x \equiv x^{45}\equiv (-2)^9 \equiv 17\bmod23$

As you have noticed, $103 (-2) \equiv 1 \bmod 23$ is definitely helpful. It let's you skip the first reduction.

lhf
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Hint $ $ It reduces to solving $\,x^{\large 5}\equiv -2\pmod{\!23}\,$ by easy congruence arithmetic. Now apply

Key Idea $ $ To take $k$'th roots when $k$ is coprime to period $n,\,$ raise to power $\,1/k\pmod{\!n},\,$ i.e.

$$\ \ \ \ \ \ \,\bbox[8px,border:2px solid #c00]{ \text{if $\:\!\ x^{\large n} =\, 1\, =\, a^{\large n} \ $ then }\ x^{\large k} = a \iff x = a^{\large (1/k)_n}}$$

so $\bmod 23\!:\ x,a\not\equiv 0\,\Rightarrow\, x^{\large 22} \equiv 1\equiv a^{\large 22} $ hence $\, x^{\large 5}\equiv a\iff x\equiv a^{\large (1/5)_{22}}\equiv a^{\large 9}\ $ by

$$\bmod 22\!:\,\ \dfrac{1}5\equiv \dfrac{3(\color{#c00}{-7})}5\equiv \dfrac{3}1\,\dfrac{\color{#c00}{15}}5\equiv 9\qquad$$

Alternatively by inverse reciprocity $$\bmod 22\!:\,\ \dfrac{1}5\equiv \dfrac{1+22\color{#c00}{(2)}}{5}\equiv 9\qquad\qquad$$

by $\bmod 5\!:\ 0\equiv 1\!+\!22\color{#c00}k\equiv 1\!+\!2k\iff 2k\equiv 4\iff \color{#c00}{k\equiv 2}$

Bill Dubuque
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