10. Find all prime numbers $p$ that satisfy the following condition: 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 $a^{2} \mid r$.
10. Solution: It is easy to verify that $p=2,3,5,7$ all satisfy the conditions. Now, let's discuss $p \geqslant 11$. If $p$ satisfies the conditions, then:
(1) $p-4$ has no prime factors greater than 4;
(2) $p-8$ has no prime factors greater than 8;
(3) $p-9$ has no prime factors greater than 9.

From (1), we get: $p-4=3^{a} (a \geqslant 2)$
From the above equation and the fact that $p-8$ is odd, we know that $p-8$ cannot be divisible by 2 or 3. Therefore, from (2), we have $p-8=5^{b} \cdot 7^{c}$
From (1) and (2), we get $5^{b} 7^{c}-3^{a}+4=0 (a \geqslant 2)$
Taking the equation modulo 3, we get $(-1)^{b}+1=0 \Rightarrow 1 \leqslant b=2 l+1, l \geqslant 0$.
Since $b \geqslant 1$, from (3) we deduce that $5 \mid 3^{a}+1 \Rightarrow 2 \leqslant a=4 m+2 \quad(m \geqslant 0)$.
From (2), we know $p-q=5^{b} \cdot 7^{c}-1$, so $p-q$ cannot be divisible by 3.
Since $b \geqslant 1$, it also cannot be divisible by 5.
Next, we prove that $c=0$.
If not, assume $c \geqslant 1$, then $7 \times p-q$. From this and (3), we know
$p-q=5^{b} \cdot 7^{c}-1=2^{d} \quad(c \geqslant 1) \Rightarrow 7 \mid 2^{d}+1$, which is impossible.
From (1), (2), and $c=0$, we get
$$
5^{b}=3^{a}-4=\left(3^{2 m+1}-2\right)\left(3^{2 m+1}+2\right) .
$$

Let $\left(3^{2 m+1}-2,3^{2 m+1}+2\right)=g$, clearly $g$ is odd and $g \mid 4$, so $g=1$.
From this, we know $3^{2 m+1}-2=1,3^{2 m+1}+2=5^{b}$, so $m=0, b=1, a=2$.
Therefore, from (1), the prime number that satisfies the condition is only 13.
In summary, the prime numbers that satisfy the conditions of this problem are: $2,3,5,7,13$.