14. * $*$ Find all prime numbers $p$, such that for all prime numbers $q<p, p-\left[\frac{p}{q}\right] q$ has no square factor.
Let $S$ be the set of all prime numbers $p$ that satisfy the given conditions. It is easy to verify that $2,3,5,7,13 \in S$. Suppose $p \in S, p>7$, then $p-4$ has no prime factor $q$ greater than 4, otherwise $p-\left[\frac{p}{q}\right] q=4=2^{2}$. Since $p-4$ is odd, $p-4=3^{a}, a \geqslant 2$. Similarly, $p-8$ and $p-9$ have no prime factors greater than 8, so $p-8=3^{a}-4=5^{b} \cdot 7^{c}$. Taking both sides modulo 3, we know $b$ is odd, and modulo 8, we know $a$ is even. If $c \neq 0$, then $p-9=5^{b} 7^{c}-1=2^{d}$ (since $p-9$ has no prime factors greater than 8 and is not divisible by $3,5,7$). Taking both sides of $5^{b} 7^{c}-1=2^{d}$ modulo 7, we find that this equation does not hold. Therefore, $c=0,3^{a}-4=5^{b}$. Hence, $5^{b}=\left(3^{\frac{a}{2}}-2\right)\left(3^{\frac{a}{2}}+2\right)$. Since $\left(3^{\frac{a}{2}}-2,3^{\frac{a}{2}}+2\right)=1$, we have $3^{\frac{a}{2}}-2=1,3^{\frac{a}{2}}+2=5^{b}$, so $a=2, b=1, p=13$. Therefore, all prime numbers $p$ that satisfy the conditions are $2,3,5,7,13$.