Exercise 2.6.4 (Solve $x^2+5x + 24 \equiv 0 \pmod {36}$)

Question

Solve $x^2+5x + 24 \equiv 0 \pmod {36}$.

Richard Ganaye · Accepted Answer

\begin{proof} Since $36 = 2^2 	imes 3^2$ and $2^2 \wedge 3^2= 1$,
$$x^2+5x + 24 \equiv 0 \pmod {36} \iff
\left\{
\begin{array}{ll}
x^2+5x + 24 \equiv 0 \pmod {2^2},\
x^2+5x + 24 \equiv 0 \pmod {3^2}.
\end{array}
ight.
$$
We solve separately these two congruences.
\begin{align*}
x^2+5x + 24 \equiv 0 \pmod {4} &\iff x(x+1) \equiv 0 \pmod 4\
&\iff x \equiv 0 \pmod 4 	ext{ or } x \equiv 3 \pmod 4,
\end{align*}
because $2 \mid x$ and $2 \mid x+1$ are incompatibel.
The second congruence gives
\begin{align*}
x^2+5x + 24 \equiv 0 \pmod {9} &\iff x^2 - 4x - 3 \equiv 0 \pmod 9\
&\iff (x-2)^2 - 7 \equiv 0 \pmod 9\
&\iff (x-2)^2 - 4^2 \equiv 0 \pmod 9\
&\iff(x-6)(x+2) \equiv 0 \pmod 9\
& \iff x \equiv 6 \pmod 9 	ext{ or } x \equiv -2 \equiv 7 \pmod 9,
\end{align*}
because $3 \mid  x-6$ and $3 \mid x+2$ are incompatible.

\bigskip
Therefore
\begin{align*}
x^2+5x + 24 \equiv 0 \pmod {36}  &\iff 
\left\{
\begin{array}{ll}
x \equiv 0 \pmod 4 	ext{ or } x \equiv 3 \pmod 4 \
x \equiv 6 \pmod 9	ext{ or } x \equiv 7 \pmod 9
\end{array}
ight.
\
&\iff \phantom{	ext{ or }}
\left\{
\begin{array}{ll}
x \equiv 0 \pmod 4  \
x \equiv 6 \pmod 9
\end{array}
ight.
	ext{ or }
\left\{
\begin{array}{ll}
x \equiv 0 \pmod 4  \
x \equiv 7 \pmod 9
\end{array}
ight.\
&\phantom{\iff} \ 	ext{ or }
\left\{
\begin{array}{ll}
x \equiv 3 \pmod 4  \
x \equiv 6 \pmod 9
\end{array}
ight.
	ext{ or }
\left\{
\begin{array}{ll}
x \equiv 3 \pmod 4  \
x \equiv 7 \pmod 9
\end{array}
ight.\
&\iff x \equiv 24, 16, 15, 7 \pmod{36}.
\end{align*}\
The solutions of $x^2+5x + 24 \equiv 0 \pmod {36}$ are $x \equiv 7,15,16,24 \pmod {36}$.
\end{proof}

Exercise 2.6.4 (Solve $x^2+5x + 24 \equiv 0 \pmod {36}$)

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Exercise 2.6.4 (Solve $x^2+5x + 24 \equiv 0 \pmod {36}$)

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