Cfrgtkky

I'm stuck with the last part of this question: solve $x^2 equiv 140 pmod{221}$.

We know that $140 = 7 times2^2times5$ and $221 = 13 times 17$.

We split the original congruence in two, so we have:

$x^2 equiv 140 pmod{13}$

$x^2 equiv 140 pmod{17}$

Applying the properties of moduli we have:

$x^2 equiv 10pmod{13} rightarrow x=pm6$

$x^2 equiv 4pmod{17} rightarrow x=pm2$

After this point, it's not clear for me how I can arrive to the complete solution. Any advice?

Now you use the Chinese Remainder Theorem. For instance, let $xinmathbb Z$ be such that $xequiv6pmod{13}$ and that $xequiv-2pmod{17}$. Since $13$ and $17$ are coprime, there are integers $alpha$ and $beta$ such that $13alpha+17beta=1$. Using the generalized Euclid algorithm, you get that, for instance, $1=4times13-3times17$. Therefore,$$8=6-(-2)=32times13-24times17.$$So,$$6-32times13=-2-24times17(=-410).$$So, take this number: $-410$. Or, better still, take $32$ ($-410equiv32pmod{221}$).

Note that
$$x^2 -140equiv x^2 -140-221=x^2−19^2=(x-19)(x+19)pmod{221}.$$
It follows that
$$xequiv pm 19equiv pm 6 pmod{13}quadtext{and}quad xequiv pm 19equiv pm 2 pmod{17}.$$ Hence we have four solutions (actually we already have two of them, i.e. $19$ and $-19$) that can be obtained by using the Chinese Remainder Theorem:
$$begin{cases}xequiv 6 pmod{13}\
xequiv 2 pmod{17}
end{cases}quad
begin{cases}xequiv 6 pmod{13}\
xequiv -2 pmod{17}
end{cases}\
begin{cases}xequiv -6 pmod{13}\
xequiv 2 pmod{17}
end{cases}quad
begin{cases}xequiv -6 pmod{13}\
xequiv -2 pmod{17}
end{cases}$$
Can you take it from here?

P.S. By the first remark, $pm 19$ are two solutions and all you need is to solve just ONE system:
$$begin{cases}xequiv 6 pmod{13}\
xequiv -2 pmod{17}
end{cases}$$
if $m$ is the solution then the fourth one is $-m$.