[Example 3.5.6] Given that $p$ is a prime number, $r$ is the remainder when $p$ is divided by 210. If $r$ is a composite number and can be expressed as the sum of two perfect squares, find $r$.

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If let $p=210 n+r, 0<\sqrt{210}$, then by $r=q m>\sqrt{210}$. $\sqrt{210}=210$, which is impossible.
Therefore, the prime factors of the composite number $r$ can only be $2,3,5,7,11,13$.
So we should classify and discuss based on the prime factor $q$ of $r$.
(1) When $q=2,3,5,7$, since
$$
p=210 n+q m,
$$

then at this time $p$ has a divisor of $2,3,5,7$, which contradicts that $p$ is a prime number.
(2) When $q=11$,
$\because$ if $r$ can be expressed as the sum of two squares, let $r=a^{2}+b^{2}$, where $a, b$ are positive integers, at this time $a^{2}+b^{2}$ can be divisible by 11.
Since $a^{2} \equiv 0,1,4,9,5,3(\bmod 11)$,
if $a^{2}+b^{2} \equiv 0 \quad(\bmod 11)$,

then $a$ and $b$ must both be divisible by 11, at this time $a^{2}+b^{2}$ can be divided by $121+121=242>210$, leading to a contradiction.
(3) When $q=13$,
since $r=q m<210$, then $m<\frac{210}{q}<17$, i.e., $m \leqslant 16$.
Also, since $m \geqslant q$, so $m=13$.
At this time $r=13^{2}=169=12^{2}+5^{2}$.
From the above, only $r=169$ meets the requirements.