<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] $$(x+2)^2 \equiv -14 \equiv 35 \pmod{49} \text{ is solvable}$$
$$\iff (-14)^{\phi(49)/2} \equiv 35^{\phi(49)/2} \equiv 1 \pmod{49} \text { (But this is false.)}$$
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 $\text{(some polynomial)} \equiv 0 \pmod{prime^k}$.
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