Let $G=\{n \in \Bbb N \mid n\leq 55 , \; \gcd(n,55)=1\}$ be the group under multiplication modulo $55$. Let $x\in G$ such that $x^2=26$ and $x>30$. What is the value of $x$?
I tried brute force and fortunately, it seems that $31$ satisfies $31^2 =26\mod 55$. But I don't know how can we solve for $x$ without hit and trial.
Modulo $55$ can be split up into a combination of modulo $11$ and modulo $5$ (by the Chinese remainder theorem). You want $x^2\equiv 26\pmod{55}$, which is equivalent to $$ \cases{x^2\equiv 26\equiv 1\pmod 5\\x^2\equiv26\equiv 4\pmod{11}} $$ There are two solutions to the top equation, and two to the lower equation. So we get in total $4$ solutions: $$ \begin{align} &x_1 \equiv \cases{1\pmod 5\\2\pmod {11}}\\ &x_2 \equiv\cases{-1\equiv 4\pmod 5\\2\pmod {11}}\\ &x_3 \equiv\cases{1\pmod 5\\-2\equiv 9\pmod {11}}\\ &x_4 \equiv\cases{4\pmod 5\\9\pmod {11}} \end{align} $$ Now we just need to check what each of those four corresponds to back in modulo $55$, and we're done. $$ x_1\equiv 46\\ x_2\equiv 24\\ x_3\equiv -x_2\equiv 31\\ x_4\equiv -x_1\equiv 9 $$
To find $x_1$ and $x_2$ I did check all numbers congruent to $2$ modulo $11$ until I found one that worked. Meaning I had to explicitly try $2, 13, 24, 35$ and $46$ and see what they reduced to modulo $5$. There are constructive ways to combine modulo $5$ and modulo $11$ back into modulo $55$ directly, but with only five candidate numbers those methods aren't worth it. Once I had $x_1$ and $x_2$, I knew the two other solutions would be $-x_1$ and $-x_2$, so I didn't need to check any more.
