Proving that $x^2 + 4$ is not divisible by $3$ I need to show the following:

For any integer $x$, $x^2 + 4$ is not divisible by $3$.  

I was trying proof by contraposition, but I do not believe that is the most efficient way to go about this.  Can anybody point me in the right direction, please?
Assume that $x^4 +4$ is divisible by $3$. Then $x^2 + 1 = (x^2 +4) -3$ is divisible by $3$.

Now (following https://math.stackexchange.com/a/630761/26188) you have that one of the following is divisible by $3$:


$x-1$
$x$
$x+1$.
If $x$ is divisible by $3$, then so is $x^2$ and so $x^2 + 1$ would not be divisible by $3$.


If $x$ is not divisible by $3$, then $x-1$ or $x+1$ would be divisible by $3$. So $(x-1)(x+1) = x^2 -1$ is divisible by $3$. And so then $x^2 + 1 = (x^2 -1) +2$ can't be divisible by $3$.
$$x \equiv 0 \equiv 1 \equiv -1 \pmod{3} \space \text{for all cases}$$

Then:

$$x^2 \equiv 0 \equiv 1 \equiv 1 \pmod{3}$$

In any of the cases, 

$$4 \equiv 1 \pmod{3}$$

$$x^2 + 4 \equiv 1 \equiv 2 \equiv 2 \pmod{3}$$

ALSO:

$$x^2 + 4 = (x + 2)^2 - 4x$$

Suppose it is divisible by $3$ then,

$$x + 2 \equiv 2\sqrt{x} \pmod{3}$$

But $\sqrt{x}$ isn't necessarily an integer, but a remainder is always an integer.