Solve the congruence by using the method of completing the square

Question

Actually, I found a open problem which is the same as my problem, see Solve quadratic congruence equation by completing square. But I can not understand the answers...they are too brief...

My typical method to solve a prime power $p^r$ modulus congruence is first solve the corresponding $p$ modulus congruence and then rise the solution in some way. For example, to solve the congruence $x^2 +x+7=0$ (mod $27)$, I will first solve the congruence $x^2 +x+7=0$ (mod $3)$ and the only solution is $1$. Then, with the same trick used in the open problem, we can rise the solution, and finally get the solutions of the congruence modulo $27$.

But my textbook requires me to solve this problem with the method of completing the square. And it provides me of a hint that $4x^2+4x+28=(2x+1)^2+27.$ However, I have no idea about this hint. And the textbook never mentions the method of completing the square.

Thanks in advance.

Answer 1

Completing the square is simply adding some amount and subtracting it in order to get partially a square. Consider for instance $x^2+2x=x^2+2x\underbrace{\color{green}{+1-1}}_{=+0}=(x\color{green}{+1})^2-1$...

Your textbook suggests that $$x^2+x+7\equiv 0\bmod{27}\stackrel{\cdot 4}{\iff}4x^2+4x+28\equiv0\bmod {27}$$ $$\iff \underbrace{(2x+1)^2}_{=4x^2+4x+1}+27\equiv(2x+1)^2+0\equiv(2x+1)^2\equiv0\bmod 27\iff \ldots$$ I obtained $$x\equiv\begin{cases}4\\13\\22\end{cases}\bmod 27$$

Answer 2

if $$ u^2 \equiv 0 \pmod {27} $$ then $$ u \equiv 0 \pmod 9 $$

so $$ u \equiv 0,9,18 \pmod {27} $$