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.
 A: $x^2 \equiv 1 \pmod{143}$ 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.
A: So we have $13\mid (x-1)(x+1)$ so $x=13k\pm 1$ and similary $x=11l\pm 1$


*

*if $13k+1 = 11l+1\implies 11\mid k\implies k=11s \implies x= 143s+1\implies \boxed{x=1}$ 

*if $13k-1 = 11l-1\implies 11\mid k\implies k=11s \implies $$x= 143s-1\implies \boxed{ x=142}$ 

*if $13k+1 = 11l-1\implies 11\mid 13k+2\implies 11\mid 2(k+1) \implies $ $k=11s-1\implies x=143s-12 \implies $ $ 
\boxed{x= 131}$ 

*if $13k-1 = 11l+1\implies 11\mid 13k-2\implies 11\mid 2(k-1) \implies $ $k=11s+1\implies x=143s+12 \implies \boxed{x= 12}$ 
A: You want to find all elements of $U_{143}$ such that $x^2\equiv1\pmod{143}$, or  equivalently, such that
$$143\qquad\text{ divides }\qquad x^2-1=(x-1)(x+1).$$
You have already found that $143=11\times13$. Do you see how to finish from here?
A: Hint:
By the Chinese remainder theorem, we have a ring isomorphism $\;\mathbf Z/143\mathbf Z\simeq \mathbf Z/11\mathbf Z\times\mathbf Z/13\mathbf Z$, so
$$U_{143}\simeq U_{11}\times U_{13}$$
so you have to solve $x^2\equiv 1\mod 11$ and $\bmod 13$, which is easy since $\mathbf Z/11\mathbf Z$ and $\mathbf Z/13\mathbf Z$ are fields, combine these solutions in all possible ways, and use the inverse isomorphism to get the solutions modulo $143$.
