Finding all elements in $x\in U_{143}$ such that $x^2=1 \pmod{143}$ 
Find all elements in $x\in U_{143}$ such that $x^2=1 \pmod{143}$

I am kind of stuck here. I know that $143=11*13$, and perhaps looking at $U_{11},U_{13}$ will help but I am unable to find a solution so far.

$x^2 \equiv 1 \pmod{143}$ is equivalent to solving the following system:
\begin{align*}
x^2 & \equiv 1 \pmod{11}\\
x^2 & \equiv 1 \pmod{13}
\end{align*}
In general for a prime $p>2$, the congruence $x^2 \equiv 1 \pmod{p}$ has only two solutions, namely $x = 1,p-1$. The reason being if $p \mid (x-1)(x+1)$ then by the prime property either $p \mid x-1$ or $p \mid x+1$. However it cannot divide both (why?).
Thus the above system reduces to the following:
\begin{align*}
x & \equiv 1 \pmod{11} & x & \equiv 1 \pmod{11} & x & \equiv -1 \pmod{11} & x & \equiv -1 \pmod{11}\\
x & \equiv 1 \pmod{13} & x & \equiv -1 \pmod{13} & x & \equiv 1 \pmod{13} & x & \equiv -1 \pmod{13}
\end{align*} 
There are quite a few ways to do this, one of them being Chinese remainder theorem.
We can solve it directly as well, for example for
\begin{align*}
x & \equiv 1 \pmod{11}\\
x & \equiv -1 \pmod{13}
\end{align*}
From the first equation we can say $x=11k+1$, then in the second equation we will have $11k+1 \equiv -1 \pmod{13}$, which is same as $11k \equiv 11 \pmod{13}$. Thus $k \equiv 1 \pmod{13}$ ($\because \, 11$ is invertible mod $13$). So we have $\color{red}{x=12}$ as a solution.
