Solve quadratic congruence equation by completing square Q: Solve the congruence $x^2+x+7\equiv 0$ (mod $27$) by using the method of completing the square from elementary algebra, thus $4x^2+4x+28=(2x+1)^2+27$. Solve this congruence (mod $81$) by the same method.
Thought:
I follow its direction to get
$$(2x+1)^2 \equiv 0 \quad \text{mod }27$$
But then  I don't know how to continue....
On the other hand, I use the method learned in class: first to solve $x^2+x+7\equiv 0$ (mod $3$), which only $x=1$ is a solution. Then it is a singular root since $f'(1)=3\equiv 0$ mod $3$. And $f(1)\equiv 0$ mod $9$ implies that $f(x)\equiv 0$ mod(9) has 3 solutions: $1, 4, 7$. Further to calculate, no solution in mod $81$, and have3 solutions in mod $27$.
But as the question required, how can I use completing square method to do? Thank you.
 A: Since I don't know your background, let me recall the definition of the $p$-adic valuation on $\mathbf Z$ associated to a prime $p$. Any $a\in \mathbf Z$ can be written uniquely as $a=a'p^n$, with $n$ maximal, and we define $v_p (a)=n$. Obviously $v_p (ab)=v_p (a)+v_p (b)$. The so called ultrametric inequality is perhaps less well known : $v_p (a+b)\ge min (v_p (a),v_p (b))$, with equality if $v_p (a)\neq v_p (b)$ (check). This is elementary but useful, because it gives us a guideline in calculations.
Back to your question: the congruence $(2x+1)^2+27\equiv 0$ mod $27$ implies $2v_3(2x+1))\ge3$, hence $v_3(2x+1)\ge \frac 32$, or $v_3(2x+1)\ge2$. But $2x+1=2(x-1)+3$, so the ultrametric inequality implies $v_3(x-1)=1$  (NB: one can see directly from the start that $v_3(x-1)\ge1$, but this is less precise). Since $\mathbf Z/27$ surjects canonically onto $\mathbf Z/27$, one checks immediately that the solutions of the congruence are $1,4,7$ mod $27$. 
The congruence $(2x+1)^2+27\equiv 0$ mod $81$ implies $v_3(LHS)\ge4$, and the ultrametric inequality requires this time that $2v_3(2x+1)=v_3(3^3)$, or $v_3(2x+1)=\frac 32$: impossible.
A: $$(2x+1)^2\equiv0\implies(2x+1)^2\equiv0,81,324\implies2x+1\equiv0,\pm9,\pm18\implies x\equiv\cdots\pmod{27}$$
