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For primes $p,q$, what are the solutions to $$x^2=(p+q)x\pmod{pq}$$

Bernard
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אבג דיזיין
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  • Please edit in your attempts to solve this problem? – Rushabh Mehta Dec 16 '20 at 13:27
  • Can't really find a way to solve this.. – אבג דיזיין Dec 16 '20 at 13:32
  • If you don't know CRT: note that it's equivalent to $,(x-p)(x-q)\equiv 0 \pmod {pq},,$ i.e. $,pq\mid(x-p)(x-q),,$ and $,pq\mid mn\iff p\mid m, q\mid n, $ or $,p\mid n,q\mid n,,$ or $,pq\mid n,$ or $,pq\mid n,,$ by unique factorization. – Bill Dubuque Dec 16 '20 at 14:29
  • And how can use it? – אבג דיזיין Dec 16 '20 at 14:34
  • Same as in the linked dupes, except you have roots $,p,q,$ vs $,-2,-3$ or $,0,1\ \ $ – Bill Dubuque Dec 16 '20 at 14:37
  • You'll get the roots $, p,\ q,\ 0,\ p+q\ \ \ $ – Bill Dubuque Dec 16 '20 at 14:44
  • Can you please upload the full answer, still can't understand what to do and how. :( – אבג דיזיין Dec 16 '20 at 14:48
  • Do you not know CRT? If you do the linked answers explain in detail how to use it to solve such quadratics. – Bill Dubuque Dec 16 '20 at 14:52
  • As an optimization note that if $,r,$ is a root of $,x(x-b),$ then so too is $,b-r,,$ so the roots come in pairs that sum to $,b,$ (as in Vieta's formula). Here $,b = p+q,$ and our root pairs are $,p,q,$ and $,0,,p+q.,$ These are the solutions of all possible combinations of the roots listed in Bernard's answer, e.g. $p$ is the root $,\equiv (p,p)\equiv (0,p)\pmod{p,q},$ and $,q,$ is that $\equiv (q,q)\equiv (q,0),$ and their sum $,p+q,$ that $,\equiv (p+q,p+q)\equiv (q,p). \ \ $ – Bill Dubuque Dec 16 '20 at 15:13
  • The $,4,$ combinations of $,x\equiv p,q\pmod {!p},,$ $,x\equiv p,q\pmod {!q},$ are $\qquad\qquad\begin{align} &(p,p), (p,q), (q,p), (q,q)!\pmod{(p,q)}^{\phantom{|^|}}\[.3em] \mapsto &\ \ \ p,\ \ \ \ \ \ \ 0,\ \ \ \ p!+!q,\ \ \ \ q\ \ \ \text{when solved by CRT}\end{align}\ \ $ – Bill Dubuque Dec 16 '20 at 15:23
  • Thank you so much! – אבג דיזיין Dec 16 '20 at 15:24
  • If anything still is not clear you can ask further questions in comments here or in the linked answers. It will be clarified when you learn CRT as a ring isomorphism $,\Bbb Z/pq\cong \Bbb Z/p\times \Bbb Z/q,$ (which I do implicitly by doing arithmetic on the CRT solution tuples). – Bill Dubuque Dec 16 '20 at 15:25

2 Answers2

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Hint:

Use the Chinese remainder theorem: for $p\ne q$, $$\mathbf Z/pq\mathbf Z\simeq\mathbf Z/p\mathbf Z\times \mathbf Z/q\mathbf Z$$ so the given congruence is equivalent to the system of congruences $$\begin{cases} x^2\equiv qx\mod p \\ x^2\equiv px\mod q \end{cases}\iff\begin{cases} x\equiv 0 &\text{or }x\equiv q\mod p \\ x\equiv 0 &\text{or } x\equiv p \mod q \end{cases}$$ Next, you can use the inverse isomorphism of the Chinese remainder theorem, which relies on a Bézout's relation $up+vq=1$: \begin{align} \mathbf Z/p\mathbf Z\times \mathbf Z/q\mathbf Z &\longrightarrow \mathbf Z/pq\mathbf Z \\ (x\bmod p,y\bmod q)&\longmapsto yup+xvq\bmod pq \end{align}

Bernard
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We can rewrite the modular equivalence as $$x^2-(p+q)x+kpq=0$$for some $k\in\mathbb Z$.

The discriminant of this quadratic is $(p+q)^2-4kpq$. To check whether there exists a square root of the discrimnant, we have to evaluate ${(p+q)^2-4kpq\choose pq}$, the legendre symbol. However, this is equal to ${(p+q)^2\choose pq}$ trivially, and this tells us that the discriminant is in fact a quadratic residue.

Hence, our solutions are of the form $$\frac{p+q\pm\sqrt{(p+q)^2-4kpq}}2$$for $k\in\mathbb Z$.

Rushabh Mehta
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