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Problem:

Find all integer values of $x$ for,

$$3^{x}\equiv 18\pmod{99}$$

Attempt:

Divide $3^{x}\equiv 18\pmod{99}$ by 9 to get

$$3^{x-2}\equiv 2\pmod{11}$$

From here, I want this modular equation to reduce to the form

$$3^{\phi(11)}\equiv 1\pmod{11}$$

by Euler's Totient Theorem, but I got stuck from there.

Are there any better approaches or elegant solutions to this problem? Any help is appreciated!

1 Answers 1

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Your solution is basically the right way to go; if you could find a single $x$ for which $3^{x-2}\equiv 2\pmod{11}$, you would then know that $x+n\varphi(11)$ was also a solution* for every integer $n\geq 0$. However, it turns out that the powers of $3$ mod $11$ are $\{1,\,3,\,9,\,5,\,4\}$, so there is actually no solution to $3^{x-2}\equiv 2\pmod{11}$, hence no solutions to the equation at all. There's not any particular way to come to this conclusion other than computing the powers of $3$ mod $11$ by hand.

*However, if you're looking for all the solutions, they would be of the form $x+nm$ where $x$ is the first integer for which it holds, $n$ is an arbitrary non-negative integer, and $m$ is the multiplicative order of $3$ mod $11$, which is $5$ rather than $10=\varphi(11)$. Again, you'd have to compute the multiplicative order, but it helps to know that $m$ divides $\varphi(11)$ for this computation. It's a moot point here, given that there are no solutions though.

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    There *are* ways besides computing all powers of $3,\,$ e.g. $3\equiv 5^2\,$ is a square hence so are all powers of $3,\,$ but $2$ is not a square (by $\,2^5\not\equiv 1)\ \ $ – 2017-01-10