0
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Chinese remainder theorem is for two different, relatively prime moduli $\endgroup$
    – J. W. Tanner
    Nov 22, 2019 at 4:29

1 Answer 1

Reset to default
2
$\begingroup$

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 ?

$\endgroup$
2
  • 1
    $\begingroup$ So one solution would be $y\equiv6\pmod7$ (and could then solve for $x$), but what if $2y+x+x^2\equiv0\pmod7$? $\endgroup$
    – J. W. Tanner
    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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .