Find all primes $p \geq 3$ with the following property: for any prime $q<p$, the number
\[
p - \Big\lfloor \frac{p}{q} \Big\rfloor q
\]
is squarefree (i.e. is not divisible by the square of a prime).
To solve the problem, we need to find all prime numbers \( p \geq 3 \) such that for any prime \( q < p \), the number 
\[ p - \left\lfloor \frac{p}{q} \right\rfloor q \]
is squarefree. This means that \( p \mod q \) must be squarefree for any prime \( q < p \).

1. **Initial Assumption and Analysis**:
   Suppose \( p > 15 \). We will analyze the properties of \( p \mod q \) for various primes \( q \).

2. **Considering \( p - 4 \)**:
   Let \( q \) be any prime factor of \( p - 4 \). If \( q > 3 \), then \( p \mod q = 4 \), which is not squarefree (since 4 is \( 2^2 \)). Therefore, \( q = 3 \). This implies that \( p - 4 \) is divisible by 3, i.e., \( p = 3^k + 4 \) for some integer \( k \).

3. **Considering \( p - 8 \)**:
   Similarly, consider \( p - 8 \). Any prime factor of \( p - 8 \) must be either 5 or 7. This implies that \( p = 3^k + 4 \) must satisfy \( p - 8 = 3^k - 4 \), which must be divisible by 5 or 7.

4. **Considering \( p - 9 \)**:
   Next, consider \( p - 9 \). Any prime factor of \( p - 9 \) must be either 2 or 7. This implies that \( p = 3^k + 4 \) must satisfy \( p - 9 = 3^k - 5 \), which must be divisible by 2 or 7.

5. **Solving the Equations**:
   We need to solve the equations \( 3^k - 5 = 2^r \) and \( 3^k - 4 = 5^r \) in natural numbers.

   - For \( 3^k - 5 = 2^r \):
     Reduce modulo 4:
     \[
     3^k \equiv 1 \pmod{4} \implies k \text{ is even}
     \]
     Let \( k = 2a \). Then:
     \[
     3^{2a} - 5 = 2^r
     \]
     Reduce modulo 3:
     \[
     (-1)^a - 5 \equiv 2^r \pmod{3} \implies r \text{ is even}
     \]
     Let \( r = 2b \). Then:
     \[
     (3^a - 2^b)(3^a + 2^b) = 5
     \]
     The only solution is \( a = 1 \) and \( b = 1 \), which gives \( k = 2 \) and \( r = 2 \). Thus, \( p = 3^2 + 4 = 13 \).

   - For \( 3^k - 4 = 5^r \):
     Reduce modulo 4:
     \[
     3^k \equiv 0 \pmod{4} \implies k \text{ is even}
     \]
     Let \( k = 2a \). Then:
     \[
     3^{2a} - 4 = 5^r
     \]
     This equation has no solutions for \( a > 1 \).

6. **Checking Small Primes**:
   We need to check if there are any primes \( p < 15 \) that satisfy the condition. The primes less than 15 are 3, 5, 7, 11, and 13.

   - For \( p = 3 \):
     There are no primes \( q < 3 \).

   - For \( p = 5 \):
     \( 5 \mod 3 = 2 \) (squarefree).

   - For \( p = 7 \):
     \( 7 \mod 3 = 1 \) (squarefree), \( 7 \mod 5 = 2 \) (squarefree).

   - For \( p = 11 \):
     \( 11 \mod 3 = 2 \) (squarefree), \( 11 \mod 5 = 1 \) (squarefree), \( 11 \mod 7 = 4 \) (not squarefree).

   - For \( p = 13 \):
     \( 13 \mod 3 = 1 \) (squarefree), \( 13 \mod 5 = 3 \) (squarefree), \( 13 \mod 7 = 6 \) (squarefree).

Thus, the primes \( p \) that satisfy the condition are \( p = 5, 7, 13 \).

The final answer is \(\boxed{5, 7, 13}\).