(11) $a, b$ are positive integers and $a b+a-b-5=0$, if $M=\max \left(a^{2}+b^{2}\right)$, then the number of integer points $(x, y)$ satisfying $3 x^{2}+2 y^{2} \leqslant M$ is $\qquad$.
51 From $a b+a-b-5=0$, we know
$$
b(a-1)+(a-1)-4=0 .
$$

If $a=1$, then $b-1-b-5=0$, which means $1-5=0$, a contradiction, hence $a \neq 1$. When $a \neq 1$, then $b+1-\frac{4}{a-1}=0$, so $b=\frac{4}{a-1}-1$. Considering $a-1=1,2,4$, i.e., $a=2, 3, 5$, we get $b=3, 1, 0$, thus the ordered pairs $(a, b)$ that satisfy the condition are $(2,3)$ and $(3, 1)$, so $M=\max (13,10)=13$. Considering the integer points on and inside the ellipse $3 x^{2}+2 y^{2}=13$, on the $y$-axis there are $(0,1),(0,-1),(0,2),(0,-2)$, a total of 4 integer points; on the $x$-axis there are $(1,0),(-1,0),(2,0),(-2,0)$, a total of 4 integer points; the origin is also an integer point that satisfies the condition; for integer points inside the ellipse but not on the axes, we only need to consider the first quadrant, and then use symmetry to find all, in the first quadrant there are $(1,1),(1,2)$ two integer points, thus in the four quadrants there are $4 \times 2=8$ (points) integer points. Therefore, on and inside the ellipse $3 x^{2}+2 y^{2}=13$ there are a total of 17 integer points.