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?

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.
