Let $p$ be a prime number such that $\frac{28^p-1}{2p^2+2p+1}$ is an integer. Find all possible values of number of divisors of $2p^2+2p+1$.
To solve the problem, we need to determine the number of divisors of \(2p^2 + 2p + 1\) given that \(\frac{28^p - 1}{2p^2 + 2p + 1}\) is an integer for a prime number \(p\).

1. **Identify the divisors of \(2p^2 + 2p + 1\):**
   Let \(q\) be a prime divisor of \(2p^2 + 2p + 1\). Since \(q\) divides \(2p^2 + 2p + 1\), it must also divide \(28^p - 1\). This is because \(\frac{28^p - 1}{2p^2 + 2p + 1}\) is an integer.

2. **Properties of \(q\):**
   Since \(q\) divides \(28^p - 1\), by Fermat's Little Theorem, we have:
   \[
   28^{q-1} \equiv 1 \pmod{q}
   \]
   This implies that the order of 28 modulo \(q\), denoted as \(\text{ord}_q(28)\), must divide both \(q-1\) and \(p\). Therefore, \(\text{ord}_q(28)\) can be either 1 or \(p\).

3. **Order of 28 modulo \(q\):**
   - If \(\text{ord}_q(28) = 1\), then \(28 \equiv 1 \pmod{q}\), which implies \(q\) divides \(27\). The prime divisors of 27 are 3. However, if \(q = 3\), then \(2p^2 + 2p + 1 \equiv 0 \pmod{3}\), which is not possible for any prime \(p\). Hence, \(\text{ord}_q(28) \neq 1\).
   - Therefore, \(\text{ord}_q(28) = p\), and \(p\) must divide \(q-1\). This implies \(q \geq p + 1\).

4. **Bounding the number of prime divisors:**
   Suppose \(2p^2 + 2p + 1\) has more than 3 distinct prime divisors. Let \(q_1, q_2, \ldots, q_k\) be the distinct prime divisors. Then:
   \[
   2p^2 + 2p + 1 \geq (p+1)^3
   \]
   However, this inequality does not hold for any prime \(p\), leading to a contradiction. Therefore, \(2p^2 + 2p + 1\) can have at most 3 distinct prime divisors.

5. **Possible cases for the number of divisors:**
   - If \(2p^2 + 2p + 1 = q\), where \(q\) is a prime, then the number of divisors is 2.
   - If \(2p^2 + 2p + 1 = q^2\), where \(q\) is a prime, then:
     \[
     2p^2 + 2p + 1 = (pk + 1)^2 = p^2k^2 + 2pk + 1
     \]
     This simplifies to:
     \[
     2p + 2 = pk^2 + 2k
     \]
     Solving for \(k\), we find that \(k = 1\) leads to a contradiction. Hence, this case is not possible.
   - If \(2p^2 + 2p + 1 = qr\), where \(q\) and \(r\) are distinct primes, then:
     \[
     2p^2 + 2p + 1 = (pk + 1)(pm + 1) = p^2mk + p(m + k) + 1
     \]
     This simplifies to:
     \[
     2p + 2 = pmk + m + k
     \]
     Since \(m\) and \(k\) are distinct, this also leads to a contradiction.

Therefore, the only possible value for the number of divisors of \(2p^2 + 2p + 1\) is 2.

The final answer is \(\boxed{2}\)