1. ## Polynomial Congruences

How do you find all the solutions of the congruence

7x^2-41x+48=0(mod333)

2. Multiplying the congruence by 28 gives $196x^2-1148x+1344\equiv0\,(\bmod{}\,333).$

Note that $\gcd(28,333)=1$ and so the second congruence implies the first as well: i.e. $7x^2-41x+48\equiv0\,(\bmod{}\,333)\ \iff\ 196x^2-1148x+1344\equiv0\,(\bmod{}\,333).$

The reason for multiplying by 28 is to make the leading coefficient a perfect square and the coefficient of $x$ even so that we can complete the square. Completing the square gives $(14x-41)^2-337\equiv0\,(\bmod{}\,333).$ Hence, letting $y=14x-41,$ we have to solve the congruence

$y^2\equiv337\,(\bmod{}\,333)\equiv4\,(\bmod{}\,333 )\quad\ldots\,\boxed1$

Clearly $y=2$ is a solution, so one solution to the original congruence is found by solving $14x-41\equiv2\,(\bmod{}\,333)$ for $x.$ I can tell you that one possiblity is $x=122$ and substituting into the original congruence shows that $x=122$ is indeed one solution.

Another solution to $\boxed1$ is $y=-2.$ Find all solutions to $\boxed1$ and for each check and see if the corresponding solution for $x$ satisfies the original congruence. Good luck.