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I am stuck on a question of Congruence . I would like u to give me a hint or a direction for what I have reached so far. I have got that: $$ (x-p_1)\cdot(x-p_2)\equiv\ 0\pmod{p_1p_2} $$ When: $p_1, p_2$ are prime.

Arturo Magidin
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    Are you asking if this is true, or is this part of a bigger question that you have not put here, and that you are asking about that bigger question? – Aiden Chow Dec 09 '20 at 23:23
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    We don't know where you are trying to go; we don't know what you are trying to say. It is impossible to give you a hint or direction. This question, right now, is not even a question. – Arturo Magidin Dec 09 '20 at 23:25
  • Are you trying to solve this equation? Expand $x^2-(p_1+p_2)x+p_1p_2\equiv0\mod{p_1p_2}$ and show it is the same as $(x-(p_1+p_2))x\equiv0\mod{p_1p_2}$. Perhaps you can see how to proceed. – robjohn Dec 10 '20 at 03:23
  • I am so sorry guys... I had to solve this equation, i.e. to find all potential results of x, when p1 and p2 are prime. I still get stuck on this. –  Dec 10 '20 at 08:15
  • See here for a general method using CRT. – Bill Dubuque Dec 10 '20 at 08:25
  • so I get: $$x\equiv\ p_1,p_2\ (mod\ p_1), and \ x\equiv\ p_1,p_2\ (mod\ p_2)? $$ –  Dec 10 '20 at 08:53
  • I forgot to tag u. @Bill Dubuque –  Dec 10 '20 at 09:58
  • Yes, now you need to lift those to up to $4$ roots mod $,p_1 p_2,$ using CRT, as in the link. You'll get the "constant" roots $,p_1,p_2,,$ and two other "mixed" roots that yield $,x\equiv 0,,p_1+p_2\ \ $ – Bill Dubuque Dec 10 '20 at 10:12

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