Example 16 (1999 National Training Team Selection Exam Question) Find all prime numbers $p$ that satisfy the following conditions: for any prime number $q<p$, if $p=k q+r, 0 \leqslant r<q$, then there does not exist an integer $a$ greater than 1 such that $a^{2}$ divides $r$.

---

The translation maintains the original text's line breaks and format as requested.
Notice that $p=2, p=3=1 \times 2+1$,
$$
p=5=2 \times 2+1=1 \times 3+2, p=7=2 \times 3+1=1 \times 5+2 .
$$

all satisfy the conditions required by the problem.
However, $p=11=1 \times 7+4=1 \times 7+2^{2}$,
$p=11$ does not satisfy the conditions required by the problem.
Now consider the case where $p>11$.
Examine $p-r=k q, 0 \leqslant r < q$.
From $p>11$ we know $r \neq 1,2,3,5,6,7,10,11,12$.
Now study $p-4, p-8, p-9$.
Since $p-4$ cannot have a prime factor greater than 3, by (1)
$p-4=3^{a}, a \geqslant 2$.
Since $p$ is an odd prime, $p-9$ is even and has a prime factor of 2, and its prime factors do not exceed 7. Since $p-9=3^{a}-5$, $p-9$ cannot have a prime factor of 5.
Thus, $p-9$ can only have prime factors of 2 and 7.
Now consider the highest power of 2 that divides $p-9$.
$p-9=3^{a}-5 \equiv\left\{\begin{array}{l}1-5 \\ 3-5\end{array} \equiv\left\{\begin{array}{l}4 \\ 6\end{array}(\bmod 8)\right.\right.$,
Thus, $8 \nmid p-9$.
Therefore, $p-9$ has only two possibilities:
$p-9=2 \cdot 7^{b}$, or $p-9=2^{2} \cdot 7^{b}$.
If $p-9=2 \cdot 7^{b}$, then $3^{a}-5=2 \cdot 7^{b}$, i.e., $3^{a}=2\left(7^{b}+1\right)+3$.
This leads to $0 \equiv 7 \equiv 1(\bmod 3)$, which is a contradiction.
Thus, only $p-9=4 \cdot 7^{b}$,
or $p-8=4 \cdot 7^{b}+1$.
Since $p-8$ does not have a prime factor greater than 7.
By $3 \mid p-4$ we get $3 \mid p-8$.
By $7 \mid p-9$ we get $7 \nmid p-8$.
Thus, $p-8$ can only have the prime factor 5,
$p-8=5^{c}$.
Therefore, from (2) and (3) we get $4 \cdot 7^{b}=5^{c}-1, b \geqslant 0, c \geqslant 1$.
If $b \geqslant 1$, consider the sequence
$\left\{5^{c}-1(\bmod 7)\right\}=\{4,3,5,1,2,0,4,3,5,1,2,0, \cdots\}$,
Thus, $c$ is a multiple of 6.
Let $c=2 c^{\prime}, c^{\prime} \geqslant 1$, then
$4 \cdot 7^{b}=25^{c^{\prime}}-1$, i.e., $24 \mid 4 \cdot 7^{b}$, which is impossible.
Therefore, $b \geqslant 1$ does not hold.
Thus, $b=0, c=1$, hence
$p-8=4 \cdot 7^{0}+1$, i.e., $p=13$.
In summary, $p=2,3,5,7,13$.