Show $x^2+y^2=9z+3$ has no integer solutions 
Show $x^2+y^2=9z+3$ has no integer solutions

So I know that $x^2+y^2=3(3z+1)$
And since $3\mid (9z+3)$ and $3\not\mid (3z+1)$ then $9z+3$ cannot be a square since it has a prime divisor which has a power less then $2$.
So does know this isn't a pythagorean triple $(x,y,\sqrt{9z+3})$ give me that there are no solutions or do I have to show more?
 A: Hint: if $n$ is an integer then $n^2\bmod 3$ is $0$ if $n$ is a multiple of $3$, and is $1$ otherwise.
In what case can it happen that $x^2+y^2$ is a multiple of $3$ then?
A: You've shown that $3|9x + 3$.  You've also shown that $9\not \mid 9x+3$.
SO that means $3|x^2 + y^2$ so but $9\not \mid x^2 + y^2$.
If $x$ and $y$ are divisible by $3$ then $x^2 + y^2$ is divisible by $9$.  So that's out.  If one of $x$ or $y$ is divisible by $3$ but the other isn't then $x^2 + y^2$ is not divisible by $3$ and that's out.
So the only options is neither $x$ nor $y$ is divisible by $3$.
So $x \equiv \pm 1 \pmod 3$ and $y \equiv \pm 1 \pmod 3$.  Which means $x^2 \equiv (\pm 1)^2 \equiv 1 \pmod 3$ and $y^2 \equiv (\pm 1)^2 \equiv 1 \pmod 3$.
So $x^2 + y^2 \equiv 2\pmod 3$ and $3\not \mid x^2 + y^2$ so that's out.
So all three options are out.
A: Since $x^2 + y^2 = 9z+3$ must be true, it must also be true that $$x^2+y^2 \equiv 3 \pmod 9$$
However, if we list out $x^2 \pmod 9$ for $0 \le x < 9$, we get $$0, 1, 4, 0, 7, 7, 0, 4, 1$$
There is no way of adding two numbers from $0, 1, 4, 7$ such that the sum is equal to $0 \pmod 9$.
Therefore, $x^2 + y^2 = 9z + 3$ has no integer solutions.
