Example 3. Find the basic Pythagorean triple, where one of the numbers is
$$
16 .
$$
Let's set the Pythagorean triplet as $x, y, z$, where $x=16$. From
$$
x=2 \times 4 \times 2=2 \times 8 \times 1
$$

and when $m=4, n=2$, $(m, n)=2 \neq 1$, so only $m=8, n=1$. At this point, $(m, n)=1$ and $m$ and $n$ are one odd and one even. Thus,
$$
\begin{array}{l}
y=m^{2}-n^{2}=8^{2}-1^{2}=63, \\
z=m^{2}+n^{2}=8^{2}+1^{2}=65 .
\end{array}
$$

Therefore, there is only one primitive Pythagorean triplet: 16, 63, 65.

For this problem, if we do not require it to be "primitive," we can also find two more sets: $\square$
One set is derived from $x=2 \times 4 \times 2$, setting $m=4, n=2$ gives
$$
y=4^{2}-2^{2}=12, z=4^{2}+2^{2}=20 \text {. }
$$

Thus, 16, 12, 20 is a Pythagorean triplet.
Another set is derived from $x^{2}=16^{2}=4 \times 32 \times 2$,

setting $m^{2}=32, n^{2}=2$, then
$$
\begin{array}{l}
y=m^{2}-n^{2}=32-2=30, \\
z=m^{2}+n^{2}=32+2=34 .
\end{array}
$$

Verification shows that $16^{2}+30^{2}=1156=34^{2}$.
Therefore, 16, 30, 34 is another Pythagorean triplet.