4

For distinct odd primes $p,q$, if $x^2\equiv a \pmod {\! p}$ is solvable and $x^2\equiv a \pmod {\!q}$ is solvable, then $x^2\equiv a \pmod {\! pq}$ is solvable.

Here, I am also assuming neither $p$ nor $q$ divides $a$.

Some students in my Elementary Number Theory class are suggesting that this is directly implied by the Chinese remainder theorem (CRT). I do not agree because I think CRT says that there is a congruence class (which we can represent by an integer, $x$, in $\{1,2,...,pq-1\}$) in which $x\equiv a \pmod {\! pq}$. The CRT does not say that this solution is a square (a quadratic residue) mod $pq$. Right?

user26486
  • 11,331
Geoffrey Critzer
  • 3,711
  • @user31415 With all due respect. Why did you edit my question?. I think it was more clear in my original version. Also I am asking a question not demanding that someone prove a theorem. – Geoffrey Critzer May 16 '15 at 21:42
  • I only changed the title and made latex more readable (I believe you don't disagree with that). Gregory Grant changed more of the content. See edit history. You can change some things if you want. I like this title a lot better though. – user26486 May 16 '15 at 21:49
  • Yes, thank you. I had not realized that you left my question in the body when I posted my comment. Also, I am a little upset right now because I apparently do not understand the full implications of the Chinese remainder theorem. – Geoffrey Critzer May 16 '15 at 21:57

2 Answers2

9

Let $ c $ be a solution of the congruence $x^2\equiv a\pmod{p}$, and let $ d $ be a solution of $ y^2 \equiv a \pmod{q}$.

By the Chinese Remainder Theorem, the system $ z \equiv c\pmod{p}$, $ z \equiv d \pmod{q} $ has a solution if p and q are coprime.

If $ z $ is a solution of the system, then $ z^2 \equiv c^2 \equiv a \pmod{p}$ and $z^2 \equiv d^2 = a \pmod{q} $. It follows that $ z^2 \equiv a \pmod{pq}$.

Andrew Au
  • 1,127
André Nicolas
  • 507,029
  • In the first line you are saying that c and d are specific solutions to x^2==a (mod p) and x^2==a (mod q). In other words, c^2==a (mod p) and d^2 ==a (mod q) Right?. In the second line you are saying that by the CRT the system x==c (mod p), x==d (mod q) has a solution modulo p*q. I agree. The third line makes no sense to me. – Geoffrey Critzer May 16 '15 at 21:50
  • If $x$ is a solution the system, then $x\equiv c\pmod{p}$ and therefore $x^2\equiv c^2\equiv a\pmod{p}$. Similarly, $x^2\equiv a\pmod{q}$. So $p$ divides $x^2-a$, and $q$ divides $x^2-a$, and therefore $pq$ divides $x^2-a$, so $x^2\equiv a\pmod{pq}$. – André Nicolas May 16 '15 at 22:03
  • OK, now I got it. Thanks. – Geoffrey Critzer May 16 '15 at 22:15
  • You are welcome. – André Nicolas May 16 '15 at 22:16
5

By CRT we can lift roots mod $\,p,q\,$ to a root mod $\,pq,\,$ for any polynomial $\,f\in\Bbb Z[x]$

Suppose that $\ f(x_p)\equiv 0\pmod p\ $ and $\ f(x_q)\equiv 0\pmod q$

By CRT there is an $\,x\equiv x_p\pmod p,\,\ x\equiv x_q\pmod q$

${\rm Thus}\quad\ {\rm mod}\ p\!:\,\ x\equiv x_p\,\Rightarrow f(x)\equiv f(x_p)\equiv 0$

${\rm and}\quad\ \ \ {\rm mod}\ q\!:\,\ x\equiv x_q\,\Rightarrow f(x)\equiv f(x_q)\equiv 0\,\ $ by the Polynomial Congruence Rule.

So $\,\ p,q\mid f(x)\,\Rightarrow\, {\rm lcm}(p,q) = pq\mid f(x),\ $ i.e. $\ f(x)\equiv 0\pmod{pq}$

The OP is the special case $\, f(x) = x^2-a$.

Community
  • 1
Bill Dubuque
  • 272,048