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Solve this equation system:

$$ \left\{ \begin{array}{c} x^2+2y+xy\equiv 3\pmod 7\\ x+2y^2+x^2y\equiv 4\pmod 7 \\ \end{array} \right. $$

I tried to turn them to linear equation to use the Chinese reminder theorem but I can't.

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    $\begingroup$ Chinese remainder theorem is for two different, relatively prime moduli $\endgroup$ Nov 22, 2019 at 4:29

1 Answer 1

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Hint : Note that when you add the equations together, you get \begin{eqnarray*} (1+y)(2y+x+x^2)\equiv 0\pmod 7.\\ \end{eqnarray*} Should be a doddle from here ?

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    $\begingroup$ So one solution would be $y\equiv6\pmod7$ (and could then solve for $x$), but what if $2y+x+x^2\equiv0\pmod7$? $\endgroup$ Nov 22, 2019 at 17:16
  • $\begingroup$ Hm... What if we use the trial-and-errors tactic to solve the rest? Like, assume x is... $\endgroup$
    – liszt16
    Dec 1, 2019 at 7:25

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