12.64 Two individuals, A and B, simultaneously solve the radical equation $\sqrt{x+a}+\sqrt{x+b}=7$. When copying the problem, A mistakenly copies it as: $\sqrt{x-a}+\sqrt{x+b}=7$, and as a result, finds one of the roots to be 12. B mistakenly copies it as $\sqrt{x+a}+\sqrt{x+d}=7$, and as a result, finds one of the roots to be 13.

Given that both individuals made no mistakes in their solving process except for the copying error, and that $a, b, d$ are all integers, find the values of $a$ and $b$.
(China Guangzhou, Chongqing, Luoyang, Fuzhou, Wuhan Junior High School Mathematics Competition, 1986)
［Solution］From the given conditions, we have

That is,
$$
\left\{\begin{array}{l}
\sqrt{12-a}+\sqrt{12+b}=7, \\
\sqrt{13+a}+\sqrt{13+d}=7 .
\end{array}\right.
$$

This can be rewritten as
$$
\left\{\begin{array}{l}
\sqrt{12-a}=7-\sqrt{12+b,} \\
\sqrt{13+a}=7-\sqrt{13+d} .
\end{array}\right.
$$
From (1), we get
$$
\begin{aligned}
12-a & =7+12+b-14 \sqrt{12+b} . \\
a & +b+7=14 \sqrt{12+b} .
\end{aligned}
$$

The left side of the above equation is an integer, so the right side must also be an integer. Therefore, $\sqrt{12+b}$ is a rational number. Since $b$ is an integer, $12+b$ must be a perfect square, which means $\sqrt{12+b}$ is an integer.
Similarly, we can prove that $\sqrt{13+d}$ is also an integer.
Let $p=7-\sqrt{12+b}, q=7-\sqrt{13+d}$, then by the given conditions, $p$ and $q$ are non-negative integers. From (1) and (2), eliminating $a$ gives
$$
p^{2}+q^{2}=25 .
$$

Solving this, we get
$$
(p, q)=(0,5),(3,4),(4,3),(5,0) .
$$

The corresponding values of $a, b$ are
$$
(a, b)=(12,37),(3,4),(-4,-3),(-13,-8) .
$$