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?
 A: Indeed, it is wrong. $49 \mid (x+2)^2+14$ implies $7\mid (x+2)^2$, and in particular $7\mid x+2$. But this means that $49\mid (x+2)^2$, and by difference $49$ should divide $14$ too, which is false.
A: 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
$$14 \equiv (x^2+4x+18)-(x^2+4x+4)\equiv 0 \pmod{49}$$
A: You are right, except the symbol $\sqrt{35}$ is not very good notation modulo $p$ because there is no way to distinguish between the two possible roots. Also, just because the number $35$ doesn't have a real square root doesn't automatically imply that it doesn't have one modulo $49$. For example, the number 7 has no real square root, but it is a square modulo $9$.
Here's how I would rewrite your proof.
Suppose that $x^2 + 4x + 18 \equiv 0$ had a solution mod $49$. Then, completing the square, we see that $(x+2)^2 \equiv 35$ has a solution mod $49$. It follows that $y^2 \equiv 35$ has a solution mod $49$. 
Therefore, it suffices to show that $35$ is a non-square mod 49. Indeed, if $y^2 \equiv 35$ then $y^4 \equiv 0$. We deduce that $7 | y$, since if $7 \nmid y$ then $49 \nmid y^4$. But then $y^2 \equiv 0$, contradiction. 
A: $$(x+2)^2 \equiv -14 \equiv 35 \pmod{49} \text{ is solvable}$$
$$\iff (-14)^{\phi(49)/2} \equiv 35^{\phi(49)/2} \equiv 1 \pmod{49} \text { (But this is false.)}$$
Theorem: If $m$ is has a primitive root (2, 4, and odd prime powers), $x^k \equiv a \pmod{m}$ has a solution if and only if $$\Large a^{\frac{\phi(m)}{gcd(\phi(m),k)}} \equiv 1 \pmod{m}$$ In that case, the equation has $gcd(k, \phi(m))$ incongruent solutions.
Another way: $(x+2)^2 \equiv 35 \pmod{49} \implies (x+2)^2 \equiv 35 \equiv 0 \pmod{7}$
$\implies (x+2)^2 \equiv 0 \pmod{49} \text{ (contradiction)}$
One more way: Use Hensel's lemma to solve for $\text{(some polynomial)} \equiv 0 \pmod{prime^k}$.
