Prove that $\sqrt{2 + 9n}$ is never an integer I'm trying to show the equation
$$x^2 \equiv 2 \mod 9$$
has no solutions, and I thought the best way might be to show that $\sqrt{2 + 9n}$ can never be an integer (for integer $n$). What might be a good way too go about this? 
I've tried a proof by contradiction similar to the proof that $\sqrt{2}$ is irrational but can't seem to find a way to arrive upon the contradiction.
 A: Suppose it was an integer. Then you have that for some $k\in\mathbb{N}$,
$$\sqrt{2+9n} = k$$
Or equivalently
$$k^2=9n+2.$$
Or
$$n = \frac{1}{9}(k^2+2).$$
Can you show that this can never hold (that is, $k^2+2$ is never divisible by $9$)?
Hint: notice this cannot work if $k$ is even so $k$ must be odd. If $k$ is odd, then $k = 2l+1$ for some integer $l$. Thus, you want to inspect what happens to $k^2+2 = 4l^2+4l+3$.
A: Hint any $x$ can be written as $9k$ or $9k+1$,or, ..., $9k+8$. Now square each of these numbers, and show that the remainder is never 2.
A: Hint $\, $ mod prime $\,p = 4n\!+\!3\!:\,\   \color{#0a0}{{-}1 \equiv x^2}\,\Rightarrow\, \color{#c00}{-1} \equiv  (\color{#0a0}{-1})^{2n+1}\! \equiv (\color{#0a0}{x^2})^{2n+1}\! \equiv x^{p-1} \equiv \color{#c00}1,\,$ by little Fermat. Hence $\,p\mid 2 = \color{#c00}{1-(-1)},\,$ contra $\,p\,$ odd. Yours is the special case $\,n=0,\,\ p=3\,\, $ which, more simply, can be verified directly: $\,{\rm mod}\ 3\!:\ x\not\equiv 0\,\Rightarrow\,x\equiv \pm1\,\Rightarrow\, x^2\equiv 1.$
A: The nice thing about modulo equations, that you can just try all the possible solutions. 
Mod 9 we have:
$$0^2\equiv1\\1^2\equiv1\\2^2\equiv4\\3^3\equiv0\\4^2\equiv7\\5^2\equiv7\\6^2\equiv0\\7^2=4\\8^2=1$$
$x$ nust be equal to one of them, And no square of them is 2.
