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The following is a system of quadratic congruences: $$\left\{\begin{array}{cl}x^{2}\equiv a&\pmod{3}\\x^{2}\equiv b&\pmod{7}\end{array}\right.$$ If $\left(\frac{a}{3}\right)=1=\left(\frac{b}{7}\right)$, where $(\ast)$ is the Legendre symbol, then above system has four solutions by the Chinese remainder theorem.

But, I'm wondering about that: Is it possible the first and second congruence has the same solution?

Since $\gcd(3,7)=1$, I guess it is impossible, but I can't explain clearly.

How do I explain it?

Give some advice. Thank you!

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    $\begingroup$ What precisely do you mean by "same solution"? $\endgroup$ – Bill Dubuque Feb 7 at 3:15
  • $\begingroup$ @BillDubuque If $x\equiv a_{1}\pmod{3}$ and $x\equiv a_{2}\pmod{7}$, same solution means an element of the set $(a_{1}+3\mathbb{Z})\cap(a_{2}+7\mathbb{Z})$. $\endgroup$ – Primavera Feb 7 at 3:22
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    $\begingroup$ I recommend that you review how CRT works here, e.g. see here. $\endgroup$ – Bill Dubuque Feb 7 at 3:39
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Let $x$ be any solution to $x^2\equiv a\pmod{3}$ and $y$ any solution to $y^2\equiv b\pmod{7}$. Then $x+3m$ is also a solution for any integer $m$, and $y+7n$ is also a solution for any integer $n$. So it suffices to show that there exist integers $m$ and $n$ such that $$x+3m=y+7n,$$ or equivalently $3m+7(-n)=y-x$. Such $m$ and $n$ exist because $3$ and $7$ are coprime. Concretely, as $$3\times(-2)+7\times(1)=1,$$ we can take $m=-2(y-x)$ and $n=-1(y-x)$ to find that $$x-6(y-x)=y-7(y-x)=7x-6y,$$ satisfies both congruences simultaneously.

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Let's assume that $\gcd(n,m)=1$ and that we have integers $\alpha,\beta \in \mathbb Z$ each solving one of the quadratic congruences such that

$$\alpha^2 \equiv a \pmod n \quad \land \quad \beta^2 \equiv b \pmod m$$

Then, using the CRT twice, we can be sure that there exist $y, z \in \mathbb Z$ satisfying

$$ \begin{align} & y \equiv \alpha \pmod n \quad \land \quad y \equiv 1 \pmod m\\ \text{and also } & z \equiv 1 \pmod n \quad \land \quad z \equiv \beta \pmod m \end{align} $$

For the product $x:=yz$, it follows that

$$ \begin{align} &x^2 \equiv y^2z^2 \equiv \alpha^21^2 \equiv a \pmod n\\ \text{and also } & x^2 \equiv y^2z^2 \equiv 1^2\beta^2 \equiv b \pmod m \end{align} $$

So, if both quadratic congruences are solvable individually and $\gcd(n,m)=1$, then both quadratic congruences are also solvable simultaneously.

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