19.1.11 * Find all positive integers $n$ such that $(n-36)(n-144)-4964$ is a perfect square.
Let $(n-36)(n-144)-4964=m^{2}, m$ be an integer, simplifying we get
$$
(n-90)^{2}-7880=m^{2}.
$$

For convenience, let $k=n-90$, then we have
$$
(k-m)(k+m)=k^{2}-m^{2}=7880=2^{3} \cdot 5 \cdot 197.
$$

Therefore,
$$
k-m=2^{a} 5^{b} 197^{c}, k+m=2^{3-a} 5^{1-b} 197^{1-c}.
$$
(Since $n=1$ does not satisfy (1), we have $n \geqslant 2$, hence $k \geqslant-88$. From (2) we know $|k| \geqslant \sqrt{7880}>88$, so it must be that $k>0$, thus $k-m, k+m$ in (3) are positive. From (3) we get
$$
k=\frac{2^{a} 5^{b} \cdot 197^{c}+2^{3-a} \cdot 5^{1-b} 197^{1-c}}{2}=2^{a-1} 5^{b} 197^{c}+2^{2-a} 5^{1-b} 197^{1-c},
$$

where $a=1$ or $2, b=0$ or $1, c=0$ or $1$. This yields $k=1971,987,399,207$, and the corresponding $n$ are $2061,1077,489,297$, which are the required values.