6. If $a, b$ are non-zero integers, and $(a, b)$ satisfies the equation $x^{2}-9 x y+y^{2}-9=0$, then $(a, b)$ is called a non-zero integer solution of the equation. Among the following statements about the non-zero integer solutions of the equation, the false statement is ( ).
(A) Non-zero integer solutions do not exist
(B) There are a finite number of non-zero integer solutions
(C) There are an infinite number of non-zero integer solutions, not in the first and third quadrants
(D) There are an infinite number of non-zero integer solutions, not in the second and fourth quadrants
6. (D).

Starting from the simplest case to find non-zero integer solutions. Clearly, when $x=0$, $y= \pm 3$ is an integer solution. Substituting $y=3$, we get the solutions $(0,3)$, $(27,3)$. Clearly, $(3,27)$ is also a solution.

Substituting $y=27$, we get two more solutions $(3,27)$, $(240, 27)$. Clearly, $(27,240)$ is also a solution. Substituting $y=240$, we get another solution. Generally, when we have a non-zero integer solution $\left(x_{n}, y_{n}\right)$ (with $y_{n}>x_{n}$), it must satisfy
$$
x_{n}^{2}-9 x_{n} y_{n}+y_{n}^{2}-9=0 \text {. }
$$

Thus, $x_{n} y_{n}>0$.
Testing, $x_{n} y_{n}0, \\
x_{n}^{\prime}=9 y_{n}-x_{n} .
\end{array}
$$

From (4), we know that $x_{n}^{\prime}$ is an integer, and from $y_{n}>x_{n}$, we know that $x_{n}^{\prime} \neq x_{n}$. Also, from (3), we know that $x_{n}^{\prime}$ is a positive integer, so $\left(x_{n}^{\prime}, y_{n}\right)$ is a new non-zero integer solution, located in the first or third quadrant. By this reasoning, the equation has infinitely many non-zero integer solutions, all located in the first or third quadrant, and not in the second or fourth quadrant. Therefore, the answer is (D).