Eg, solve $x^{11} \equiv 7 \pmod{61}$
The solution is $x \equiv 31 \pmod{61}$. But how do I get there? Brute checking every #, from $0 \to 60$ is very difficult?
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Hint $\ {\rm mod}\ 60\!:\ \color{#c00}{11^2\equiv 1},\ $ so $\,\ {\rm mod}\ 61\!:\ \left[\, x^{\large \color{#c00}{11}}\equiv 7\,\right ]^{\large \color{#c00}{11}}\!\Rightarrow\, x\equiv 7^{\large 11}\!\equiv 31$
Remark $\ $ If you are familar with modular fractions then the above can be viewed more intutively as simply raising $\,x^{11}\equiv 7\pmod{61}\,$ to the power $\,\frac{1}{11}.\,$ By Fermat, the exponents are determined mod $60,\,$ and $\,\dfrac{1}{11}\equiv 11\pmod{60},\,$ which yields said solution.
Bill Dubuque
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Very nice. I guess in more general terms, $\color{red}{11}\cdot \color{blue}{11} \equiv 1 \bmod 60$, so $\bmod 61$, $x^\color{red}{11} \equiv 7 \implies x\equiv 7^ \color{blue}{11} $ – Joffan Nov 02 '16 at 01:21
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1@Joffin yes, if $,AB\equiv 1\pmod{p-1},$ then $, g(x) = x^B,$ is the inverse of $,f(x) = x^A\pmod p\ $ – Bill Dubuque Nov 02 '16 at 01:23
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Which implies the converse holds (so no need to verify that the candidate solution is a valid solution). – Bill Dubuque Nov 02 '16 at 01:28
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Right, and the solution only exists if the exponent has an inverse in the appropriate reduced modulo (?) – Joffan Nov 02 '16 at 01:32
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@Joffan The method requires invertibility - see my Remark above. – Bill Dubuque Nov 02 '16 at 01:36
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See here for more on this method. – Bill Dubuque Mar 15 '22 at 17:23
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All equivalences $\bmod 61$
\begin{align} x^{11} &\equiv 7 \\ x^{22} &\equiv 49 \equiv -12 \\ x^{44} &\equiv 144 \equiv 22 \\ x^{66} &\equiv -264 \equiv -20 \quad \equiv x^{6} \text{ by Fermat's Little Theorem}\\ x^{12} &\equiv 400 \equiv 34 \\ x &\equiv 34 \cdot 7^{-1} \equiv 34\cdot 35 \equiv 1190 \equiv 31 \end{align}
The only involved part there being finding $7^{-1} \equiv 35 \bmod 61$, which I quickly got by observing $7\cdot9 \equiv 2 $ and so $7\cdot 36 \equiv 8 $ .
Check: $31^2 \equiv 961 \equiv 46 \equiv -15 \\ 31^4 \equiv 225 \equiv 42 \equiv -19\\ 31^5 \equiv -589 \equiv 21 \\ 31^{10} \equiv 441 \equiv 14 \\ 31^{11} \equiv 434 \equiv 7 \quad \checkmark$
Joffan
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