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I am trying to prove that, where $2^x = 11$ mod $61$, that ant solution $x$ can be divided by 5. I used Pohlig-Hellman to get to an equation such that $2^{12x} = 1$ mod $61$ by setting my q as 5 s.t. $A = 2^{\frac{61 - 1}{5}}$ and $B = 2^{\frac{61 - 1}{5}}$. The only solution to this is $ x = 0 \mod 5$.

Does the fact that the only solution in $\mod 5$ is $0$ show that x is divisible by 5?

Arctic Char
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user3471031
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    That's definition of what it means to be $0$ modulo $5$. Rather, $n\equiv m\pmod 5$ if $5$ divides $n-m$. – pancini Oct 03 '20 at 00:57

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Since you know $\,2\,$ generates $\,\Bbb Z_{61}^*$ if follows that $\,2\,$ has order $\color{#c00}{60}$ modulo $61$. Therefore, by the Order Theorem we deduce that $\bmod 60\!:\ 2^{\large 12x}\equiv 1\!\iff \color{#c00}{60}\mid 12x\smash{\underset{\large \div\, 12\!}\iff}\, 5\mid x\!\iff x\equiv 0\pmod{\!5}$

since, by definition, $\ a\equiv b\pmod{\!n}\iff n\mid a-b,\,$ i.e. $\,n\,$ divides $\,a-b$

Bill Dubuque
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