12. (10 points) A certain natural number minus 39 is a perfect square, and minus 144 is also a perfect square. Find this natural number.


【Analysis】Let this natural number minus 39 be $a^{2}$, and minus 144 be $b^{2}$. According to the problem, we have $a^{2}+39=b^{2}+144$, which simplifies to: $a^{2}-b^{2}=105, (a+b)(a-b)=105$. Factorizing 105 into prime factors, $105=3 \times 5 \times 7$. Then, combining the factors 3, 5, and 7, we get four corresponding values for $(a+b)$ and $(a-b)$. We can then solve the system of equations further.

【Solution】Solution: Let this natural number minus 39 be $a^{2}$, and minus 144 be $b^{2}$. According to the problem, we have:
$$
a^{2}+39=b^{2}+144,
$$

which means, $a^{2}-b^{2}=105$,
$$
\begin{array}{l}
(a+b)(a-b)=105, \\
105=3 \times 5 \times 7,
\end{array}
$$

Therefore, $(a+b)$ and $(a-b)$ have four corresponding values:
$$
\left\{\begin{array}{l}
a+b=15 \\
a-b=7
\end{array},\left\{\begin{array}{l}
a+b=21 \\
a-b=5
\end{array},\left\{\begin{array}{l}
a+b=35 \\
a-b=3
\end{array},\left\{\begin{array}{l}
a+b=105 \\
a-b=1
\end{array}\right. \text {, }\right.\right.\right.
$$

The corresponding four sets of solutions that meet the problem's requirements are:
$$
\left\{\begin{array}{l}
a=11 \\
b=4
\end{array},\left\{\begin{array}{l}
a=13 \\
b=8
\end{array},\left\{\begin{array}{l}
a=19 \\
b=16
\end{array},\left\{\begin{array}{l}
a=53 \\
b=52
\end{array},\right.\right.\right.\right.
$$

Therefore, this natural number can be:
$$
\begin{array}{c}
11^{2}+39=160, \\
13^{2}+39=208, \\
19^{2}+39=400, \\
53^{2}+39=2848,
\end{array}
$$

Answer: This natural number can be $160, 208, 400, 2848$.