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I found this sentence in a paper :
Let $n=pq$ where $p, q$ are primes and let $Qr(n) = \{x^2 \mod n | \; x \in Z_n^*\}$ the set of quadratic residues in $Z^*_n$. By Chinese remainder theorem we get :$$x \in Qr(n) \iff x \in Qr(p) \wedge x \in Qr( q)$$
It is not obvious to me why it holds. This is my effort to prove the first direction:
$x \in Q(n) \implies \exists a \in Z_n^*:\; x = a^2 \mod n$
Now I am trying to use something like inverse chinese remainder:
If $x \in Qr(p) \implies \exists b \in Z_p\; : x = b^2 \mod p$
and if $x \in Qr(p) \implies \exists b' \in Z_q\; : x = b'^2 \mod q$
Applying CRT:
$$x = b^2\frac{p}{pq}N_1 + b'^2\frac{q}{pq}N_2$$ where $N_1\frac{1}{q} = 1 \mod p \wedge N_2 \frac{1}{p} = 1\mod q$
I am not sure how to continue.

Bill Dubuque
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tonythestark
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  • Apply the Remark in the linked dupe to the polynomial $,f(x) = x^2 - k\ \ $ – Bill Dubuque Dec 19 '22 at 12:17
  • I checked the link . Do you suggest considering $f(x) = x^2 -a$ and solving it in $Z_n$ which would be the same as solving $f(x) = 0$ in $Z_p$ and $Z_q$ ? But why is that? This looks like an inverse chinese remainder – tonythestark Dec 19 '22 at 13:37
  • As in the link, suppose $a$ is a square mod $p$ & $q$, so $,a\equiv r^2\pmod{!p},, a\equiv s^2\pmod{!q}$. By CRT there is an integer $x$ with $,x\equiv r\pmod{!p}$ & $,x\equiv s\pmod{!q},$ so $,x^2\equiv r^2\equiv a\pmod{!p},,$ and $,x^2\equiv r^2\equiv a\pmod{!p},,$ so $,x^2\equiv a\pmod{!p\ &\ q}\Rightarrow x^2\equiv a\pmod{!pq}$ by $p,q$ coprime & CRT, therefore $,a,$ is a square mod $pq\ \ $ – Bill Dubuque Dec 19 '22 at 14:59

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