I am trying to find $x$ such that $x\equiv23^{\frac{1}{7}}\mod 143$, but am not really sure where to start.
I would expect that, in order to solve this, I would have to rewrite my formula to something as follows:
$$ x\equiv 23^{\frac{1}{7}}\mod 143 $$ $$ \frac{1}{23^{7}}x\cdot \equiv 1\mod 143$$
This still, however, doesn't bring me any further. Am I using the correct appraoch or should I look at it from a different angle?
You need to find which $n\in\{0,1,2,\ldots,142\}$ satisfy the equation $$ n^7=23\mod 143. $$ Hint. Since $143=11\cdot 13$, then start with solving $$ n^7=1\mod 11\quad\text{and}\quad n^7=10\mod 13 $$
Start with $143 = 11 \times 13$. You want to solve $x^7 \equiv 23 \mod 11$ and $y^7 \equiv 23 \mod 13$, and combine the two using Chinese Remainder Theorem.
This is a RSA problem for $143=pq$ with $p=11, q=13$ because $7$ is relative prime to $\phi(pq)=(p-1)(q-1)=120.$
To get the solution for $x^e \equiv a \bmod (pq),$ compute $d \equiv e^{-1}\bmod{\phi(pq)}$ and then $x\equiv a^d \bmod {pq}.$
In your case $d \equiv 7^{-1}\equiv 103 \bmod{120}$ and $x\equiv 23^{103} \equiv 23\bmod {143}.$
Check that $23^7 \equiv 23 \bmod {143}.$
We need $x^7\equiv23\pmod{11}\equiv1\ \ \ \ (1)$
Clearly, $(x,11)=1\implies x^{11-1}\equiv1\pmod{11}$
$x^1=x^{21-20}=(x^7)^3(x^{10})^2\equiv1^3\cdot1^{-2}\equiv1\ \ \ \ (1A)$
and $x^7\equiv23\pmod{13}\equiv-3\ \ \ \ (2)$
We have $x^{13-1}\equiv1\pmod{13}$
$x^1=x^{49-48}=(x^{12})^{-4}(x^7)^7\equiv1^{-4}\cdot(-3)^7\equiv-3\ \ \ \ (2A)$ as $(-3)^3\equiv-1\pmod{13},(-3)^6\equiv?$
Apply CRT on $(1A),(2A)$
