<s>[INST] Show $x^2+4x+18\equiv 0\pmod{49}$ has no solution My method was to just complete the square:
$x^2+4x+18\equiv 0\pmod{49}$
$(x+2)^2\equiv -14\pmod{49}$
$x+2\equiv \sqrt{35}\pmod{49}$
So $x\equiv\sqrt{35}-2\pmod{49}$, which has no real solutions.

I feel that this may be too elementary, is this the correct way to solve this?
 [/INST] This isn't elegant, but it doesn't take that long. Suppose, for contradiction, that there is a solution modulo $49$. In order for there to be a solution modulo $49$, there must be a solution modulo $7$. 
$$x^2+4x+4\equiv \begin{cases}
4  \pmod{7} &\mbox{if } x \equiv 0 \pmod{7}\\
2  \pmod{7} &\mbox{if } x \equiv 1 \pmod{7}\\
2  \pmod{7} &\mbox{if } x \equiv 2 \pmod{7}\\
4  \pmod{7} &\mbox{if } x \equiv 3 \pmod{7}\\
1  \pmod{7} &\mbox{if } x \equiv 4 \pmod{7}\\
0  \pmod{7} &\mbox{if } x \equiv 5 \pmod{7}\\
1  \pmod{7} &\mbox{if } x \equiv 6 \pmod{7}\\
\end{cases}
$$
Now, following Paolo Leonetti, if $x \equiv 5 \equiv -2 \pmod{7}$, then $$x^2+4x+4 \equiv 0 \pmod{49}$$
But this, along with our original equivalence, gives the ridiculous conclusion
$$14 \equiv (x^2+4x+18)-(x^2+4x+4)\equiv 0 \pmod{49}$$
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