Hi'
I need help in solving the equation:
x^25=2 (mod 133)
Thank's in advance.
Suppose $\displaystyle x \equiv 3 \pmod{19}$ and $\displaystyle x \equiv 2 \pmod{7}$. Then, by the Chinese Remainder Theorem, $\displaystyle x \equiv 79 \pmod{133}$. Indeed, $\displaystyle 79^{25} \equiv 2 \pmod{133}$. This is a solution (so the solutions 14 and 3 of the second equation are not incorrect... at least 3 is correct).
Suppose $\displaystyle x \equiv 2 \pmod{19}, x\equiv 2 \pmod{7}$. Then, $\displaystyle x \equiv 2\pmod{133}$, and $\displaystyle 2^{25} \equiv 128 \pmod{133}$, so this is not a solution.
Suppose $\displaystyle x \equiv 14 \pmod{19}, x\equiv 2\pmod{7}$. Then, $\displaystyle x \equiv 128 \pmod{133}$, and $\displaystyle 128^{25} \equiv 79 \pmod{133}$, so again, this is not a solution.
Hence, the only solution is $\displaystyle x \equiv 79 \pmod{133}$.
Use an online Chinese Remainder Theorem Calculator? You need familiarity with the subject to be able to calculate solutions otherwise.
Here is a link to a calculator that will do it for you: Chinese Remainder Theorem Calculator
How it works: Suppose $\displaystyle x \equiv 3 \pmod{19}, x \equiv 2 \pmod{7}$
Then, we know $\displaystyle x = 3+19a$ for some integer $\displaystyle a$. So, the possible equivalence classes of $\displaystyle x \pmod{133}$ are $\displaystyle 3, 22, 41, 60, 79, 98, 117$. We check each $\displaystyle \pmod{7}$ and discover that only $\displaystyle 79 \equiv 2 \pmod{7}$.
Do the same for $\displaystyle x \equiv 2 \pmod{19}$ or $\displaystyle x \equiv 14 \pmod{19}$.