On 6. Person A and Person B simultaneously solve the radical equation $\sqrt{x+a}+$ $\sqrt{x+b}=7$. When copying, A mistakenly copies it as $\sqrt{x-a}+\sqrt{x+b}=7$, and ends up solving one of its roots as 12. B mistakenly copies it as $\sqrt{x+a}+\sqrt{x+d}$ $=7$, and ends up solving one of its roots as 13. It is known that apart from the copying errors, both have correct solving processes, and $a, b, d$ are all integers. Try to find the values of $a$ and $b$. (1986, Guangzhou, Fuzhou, Chongqing, Wuhan Four Cities Competition)
Given that one root of $\sqrt{x-a}+\sqrt{x+b}=7$ is 12, we have
$$
\sqrt{12-a}+\sqrt{12+b}=7.
$$

Also, given that one root of $\sqrt{x+a}+\sqrt{x+d}=7$ is 13, we have
$$
\sqrt{13+a}+\sqrt{13+d}=7.
$$

According to the definition of arithmetic roots, we have $12-a \geqslant 0, 13+a \geqslant 0$.
Thus, $0 \leqslant 12-a \leqslant 25, 0 \leqslant 13+a \leqslant 25$.
From (1), we get $\sqrt{12+b}=7-\sqrt{12-a}$.
Squaring both sides, we get
$$
\begin{aligned}
12+b & =49-14 \sqrt{12-a}+(12-a) \\
& =81-14 \sqrt{12-a}.
\end{aligned}
$$

When $a$ takes any integer value such that $a \leqslant 12$, $\sqrt{12-a}$ is either an integer or an irrational number. From (4), we know that $\sqrt{12-a}$ cannot be an irrational number, so $\sqrt{12-a}$ must be an integer, meaning $12-a$ is a perfect square.
Similarly, $13+a$ is also a perfect square.
Using the inequality (3), $a$ can only take the values
$$
\begin{array}{l}
12,11,8,3,-4,-13, \\
-13,-12,-9,-4,3,12.
\end{array}
$$

The common values of these two sets of numbers are $12, 3, -4, -13$.
Substituting these $a$ values into (1) and (2), we get the corresponding integer values of $b$ and $d$ as follows:
$$
\left\{\begin{array}{l}a=12, \\ b=37, \\ d=-9,\end{array}\left\{\begin{array}{l}a=3, \\ b=4, \\ d=-4,\end{array}\left\{\begin{array}{l}a=-4, \\ b=-3, \\ d=-3,\end{array}\left\{\begin{array}{l}a=-13, \\ b=-8, \\ d=36 .\end{array}\right.\right.\right.\right.
$$
Therefore, the values of $a$ and $b$ are
$$
\left\{\begin{array}{l}
a=12, \\
b=37;
\end{array} \left\{\begin{array}{l}
a=3, \\
b=4;
\end{array} \left\{\begin{array}{l}
a=-4, \\
b=-3;
\end{array} \left\{\begin{array}{l}
a=-13, \\
b=-8.
\end{array}\right.\right.\right.\right.
$$

This is a problem of finding integer solutions to irrational equations, which can be transformed into finding integer solutions to a system of linear equations. The key to solving this problem lies in deducing from the given equations $\sqrt{12-a}+\sqrt{12+b}=7$ and $\sqrt{13+a}+\sqrt{13+d}=7$ that $\sqrt{12-a}$ and $\sqrt{13+a}$ are integers, and thus $12-a$ and $13+a$ are perfect squares. To reach this conclusion, the following fact is used: if $c$ is an integer and $c \geqslant 0$, then $\sqrt{c}$ is either an integer or an irrational number. This fact can be proven similarly to the proof that $\sqrt{2}$ is irrational. Students interested in this can try to prove it themselves.

In the solution process, inequalities are used to derive that from $-13 \leqslant a \leqslant 12$, we get $0 \leqslant 12-a \leqslant 25, 0 \leqslant 13+a \leqslant 25$. Therefore, this method of finding integer solutions to indeterminate equations is named the "inequality method" or "estimation method".