Once upon a math class, a student incorrectly performed the squaring of $(a+2 b-3)^{2}$, and the result was $a^{2}+$ $4 b^{2}-9$. At the teacher's request, to verify, the student substituted a natural number for $a$ and $b$. After the substitution, the result turned out to be correct. Which numbers could the student have substituted?
Solution. Let's compare the two results:

$$
a^{2}+4 a b+4 b^{2}-6 a-12 b+9=a^{2}+4 b^{2}-9
$$

After simplification, we get

$$
4 a b-6 a-12 b+18=0
$$

Find the values of $a$ and $b$ (integers) for which the equality holds.

Transform into a product:

$$
2 a(2 b-3)-6(2 b-3)=(2 a-6)(2 b-3)=0
$$

The second factor is odd for integer $b$, so it cannot be 0, hence only $2 a-6=0$ is possible. From this, $a=3$. Substitute into (1):

$$
12 b-18-12 b+18=0
$$

and this does not depend on the value of $b$.

Thus, the student substituted 3 for $a$. It is not possible to determine what value was substituted for $b$.