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Let $p,q$ be prime numbers , find all the solutions to the congruence $x^2\equiv (p+q)x \pmod{pq}$

Attempt:

$$x^2\equiv (p+q)x \pmod{pq} \Longrightarrow \cases {x^2\equiv (p+q)x \pmod{p}\\x^2\equiv (p+q)x \pmod{q}} $$

For $\pmod{p}$

$$x^2\equiv (p+q)x \pmod{p} \Longrightarrow x(x-(p+q))\equiv 0 \pmod{p}$$

Because $p$ is prime number then we can write :

$$x\equiv 0 \pmod{p}\space\space\space(1)$$ and $$x\equiv (p+q) \pmod{p}\Longrightarrow x\equiv q \pmod{p}\space\space\space(2)$$

For $\pmod{q}$

$$x^2\equiv (p+q)x \pmod{q} \Longrightarrow x(x-(p+q))\equiv 0 \pmod{q}$$

Because $q$ is prime number then we can write :

$$x\equiv 0 \pmod{q}\space\space\space(3)$$ and $$x\equiv (p+q) \pmod{q}\Longrightarrow x\equiv p \pmod{q}\space\space\space(4)$$

Now, we can say because $(1)$ and $(3)$ that $x\equiv 0 \pmod{pq}$.

There are more solutions using $(2)$ and $(4)$ but I cannot see a legal way to solve them.

Edit: note that $p,q$ can(not must) be equal to one another, which I think is another case for a solution.

Bill Dubuque
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Isaac S
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1 Answers1

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Your ‘and’ should be ‘or’.

Hint:

You just have to use the inverse isomorphism of the isomorphism of the Chinese remainder theorem, which relies on a Bézout's identity: $\enspace up+vq=1$, between $p$ and $q:$

The system of congruences $\begin{cases}x\equiv a\mod p,\\x\equiv b\mod q,\end{cases}$ has as solution $\enspace x\equiv b up+a vq\mod pq.$

Bernard
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