Let $p_1,p_2,p_3$ be primes with $p_2\neq p_3$ such that $4+p_1p_2$ and $4+p_1p_3$ are perfect squares. Find all possible values of $p_1,p_2,p_3$.
1. Let \( 4 + p_1 p_2 = x^2 \) and \( 4 + p_1 p_3 = y^2 \). This implies:
   \[
   p_1 p_2 = x^2 - 4 = (x-2)(x+2)
   \]
   and
   \[
   p_1 p_3 = y^2 - 4 = (y-2)(y+2)
   \]

2. Since \( p_1, p_2, p_3 \) are primes, and \( p_2 \neq p_3 \), we need to consider the factorization of \( (x-2)(x+2) \) and \( (y-2)(y+2) \) into products of primes.

3. Without loss of generality, assume \( p_2 > p_3 \). We need to find values of \( x \) and \( y \) such that both \( (x-2)(x+2) \) and \( (y-2)(y+2) \) are products of primes.

4. Consider the possible values for \( x \) and \( y \):
   - If \( p_1 = x+2 \), then \( p_2 = x-2 \).
   - If \( p_1 = x-2 \), then \( p_2 = x+2 \).

5. Similarly, for \( y \):
   - If \( p_1 = y+2 \), then \( p_3 = y-2 \).
   - If \( p_1 = y-2 \), then \( p_3 = y+2 \).

6. We need to find all triplets \( (p_1, p_2, p_3) \) such that each element is a prime number. Consider the triplet \( (p, p+4, p-4) \) and check for primality.

7. Taking modulo 3, we observe that one element of the triplet must be 3 for it to be prime. Therefore, let \( p_3 = 3 \).

8. If \( p_3 = 3 \), then:
   - \( 4 + p_1 \cdot 3 = y^2 \)
   - \( p_1 \cdot 3 = y^2 - 4 \)
   - \( p_1 = \frac{y^2 - 4}{3} \)

9. We need \( \frac{y^2 - 4}{3} \) to be a prime number. Check for small values of \( y \):
   - If \( y = 5 \), then \( p_1 = \frac{5^2 - 4}{3} = \frac{25 - 4}{3} = 7 \), which is a prime.
   - Therefore, \( p_1 = 7 \).

10. Now, check for \( p_2 \):
    - \( 4 + 7 \cdot p_2 = x^2 \)
    - \( 7 \cdot p_2 = x^2 - 4 \)
    - \( p_2 = \frac{x^2 - 4}{7} \)

11. We need \( \frac{x^2 - 4}{7} \) to be a prime number. Check for small values of \( x \):
    - If \( x = 11 \), then \( p_2 = \frac{11^2 - 4}{7} = \frac{121 - 4}{7} = 17 \), which is a prime.
    - Therefore, \( p_2 = 11 \).

12. Thus, the only triplet that satisfies the conditions is \( (7, 11, 3) \).

13. Since \( p_2 \neq p_3 \), the other possible triplet is \( (7, 3, 11) \).

The final answer is \( \boxed{ (7, 11, 3) } \) and \( (7, 3, 11) \).