How to solve $2^{53x} \equiv 3 \pmod{23}$?

Question

I tried to solve this congruence, but I don't understand how to find “?”

$2^{11} \equiv 1 \pmod{23}$ so

$53x \equiv ~? \pmod{11}$

I know that “?” is equal to $8$, but I don't know how to find it, thank you!

Answer 1

Finding the ???, or equivalently finding $y$ such that $2^y\equiv3\pmod{23}$, is an instance of a discrete logarithm problem. There is no universal efficient method for solving it. Thankfully $11$ is a small number, so you can brute force it.

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If you want to see a trick then an ad hoc answer can be derived from the observation that $$ 3^3=27\equiv 2^2\pmod{23}. $$ Leaving that as a hint :-)

Plug in $3=2^y$, and solve $y$ from the resulting congruence modulo $11$.

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Observations like the one above are the basis of an algorithm called index calculus. For a (barely) pen-and-pencil-doable example of index calculus you may take a look at this answer of mine, partly prepared for referrals like this. That example works in the field $\Bbb{F}_{2^{10}}$. Meaning that the arithmetic is done modulo a polynomial (with coefficients modulo two) instead of modulo some prime $p$ like here. The principles are the same.

Answer 2

$$2^8\equiv 3 \mod23$$

$$2^{11}\equiv 1 \mod23$$

$$2^{11t}\times2^{8}\equiv 3 \mod23$$

$$2^{53x}\equiv 2^{11t+8}\equiv 3 \mod23$$

$$53x\equiv 8 \mod11$$

$$9x\equiv 8 \mod11$$

$$x\equiv 7 \mod11$$

$$x=7+11k$$

where $k \in Z$