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QUESTION
Decide whether or not the following quadratic equations have
solutions. If there are any solutions, find them.
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\item[(i)]
$3x^2+5x+1\equiv0$ mod 7.
\item[(ii)]
$2x^2+3x+6\equiv0$ mod 11.
\end{description}
ANSWER
\begin{description}
\item[(i)]
A quadratic congruence $ax^2+bx+c\equiv0$ mod $p$, where $p$ is an
odd prime such that $p\not|a$, has roots if and only iff the
discriminant $d=b^2-4ac$ is a square mod $p$.
Here $d\equiv25-4.3.1\equiv13\equiv-1$ mod 7. By Theorem 7.1(v),
$-1$ is not a square mod 7, so our equation has no solutions.
\item[(ii)]
$2x^2+3x+6\equiv0$ mod 11.
Arguing as in (i), $d\equiv9-2.4.5\equiv-39\equiv5$ mod 11. Now
$5\equiv 16\equiv4^2$ mod 11, so here solutions exist.
By multiplying our original equation through by 8 $(=4a)$ we
obtain $(4x+3)^2\equiv 9-8.6\equiv5$ mod 11, so on noting that
$5\equiv (\pm4)^2$ mod 11 we see that we have two possible
solutions, viz. $4x+3\equiv4$ mod 11 and $4x+3\equiv-4$ mod 11.
These simplify to $4x\equiv1$ mod 11 and $4x\equiv-7\equiv4$ mod
11, giving solutions $x\equiv 3$ and $x\equiv 1$ mod 11,
respectively.
[Other methods of solution are possible - e.g. you may have
spotted 1 as a root, and then factorised the equation as
$(x-1)(2x-6)\equiv0$ mod 11.]
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