BdMO Online Forum

Find $x$ such that $x^2 - 3x + 7 \equiv 2 (mod 11)$

Check my solution:
This equation can be rewritten :
$x(x-3) \equiv -5 \equiv 6 (mod 11)$
But if we check the factors of $6$,we will find
that there is no solution of $x$.

$x = 11n + 7$ for any non-negative integer n.

This can be proved very easily. $x \equiv {0,1,2,3,4,5,6,7,8,9,10}$. Now checking the values for $x(x-3) mod 11$ gives the only solution when $x \equiv 7(mod 11)$.

I missed the point.Thanks to dhasan.
Correct solution:
This equation can be rewritten :
$x(x-3) \equiv -5 \equiv 6 (mod 11)$
If we check the factors of $6$,we will find
that the solution is $x \equiv 7$.

x(x-3) = -5 = 6 (mod 11)
by testing the factors of 6 we get that x = 7 (mod 11)