$ a \minus{} )$ Find all prime $ p$ such that $ \dfrac{7^{p \minus{} 1} \minus{} 1}{p}$  is a perfect square

$ b \minus{} )$ Find all prime $ p$ such that $ \dfrac{11^{p \minus{} 1} \minus{} 1}{p}$  is a perfect square
### Part (a)
We need to find all prime \( p \) such that \( \frac{7^{p-1} - 1}{p} \) is a perfect square.

1. **Initial Setup and GCD Property**:
   Since \(\gcd(7^{\frac{p-1}{2}} - 1, 7^{\frac{p-1}{2}} + 1) = 2\) for any odd prime \( p \), if \( \frac{7^{p-1} - 1}{p} = x^2 \) for some integer \( x \), then we have two cases to consider:
   - Case 1: \( 7^{\frac{p-1}{2}} - 1 = 2py^2 \) and \( 7^{\frac{p-1}{2}} + 1 = 2z^2 \)
   - Case 2: \( 7^{\frac{p-1}{2}} - 2 = 2y^2 \) and \( 7^{\frac{p-1}{2}} + 1 = 2pz^2 \)

2. **Case 1 Analysis**:
   - From \( 7^{\frac{p-1}{2}} - 1 = 2py^2 \), we have \( 6 \mid 7^{\frac{p-1}{2}} - 1 \). This implies \( 3 \mid py^2 \).
   - If \( p = 3 \), then \( \frac{7^2 - 1}{3} = 16 = 4^2 \), which is a perfect square.
   - If \( p \geq 5 \), then \( 3 \mid y^2 \) implies \( 9 \mid 7^{\frac{p-1}{2}} - 1 \). Thus, \( 3 \mid \frac{p-1}{2} \). Let \( k = \frac{p-1}{6} \). Then \( 2z^2 = 7^{3k} + 1 = (7^k + 1)(7^{2k} - 7^k + 1) \).
   - For \( 7^{2k} - 7^k + 1 \) to be a perfect square, we need \( (7^k - 1)^2 \leq 7^{2k} - 7^k + 1 \leq (7^k)^2 \), which is not possible.

3. **Case 2 Analysis**:
   - Since 2 is a quadratic residue modulo 7, but \(-1\) is not, this case does not yield any valid solutions.

4. **Conclusion for Part (a)**:
   Therefore, the only prime \( p \) for which \( \frac{7^{p-1} - 1}{p} \) is a perfect square is \( p = 3 \).

### Part (b)
We need to find all prime \( p \) such that \( \frac{11^{p-1} - 1}{p} \) is a perfect square.

1. **Initial Setup and GCD Property**:
   Since \(\gcd(11^{\frac{p-1}{2}} - 1, 11^{\frac{p-1}{2}} + 1) = 2\) for any odd prime \( p \), if \( \frac{11^{p-1} - 1}{p} = x^2 \) for some integer \( x \), then we have:
   - Case 2: \( 11^{\frac{p-1}{2}} - 2 = 2y^2 \) and \( 11^{\frac{p-1}{2}} + 1 = 2pz^2 \)

2. **Case 2 Analysis**:
   - Since 2 is not a quadratic residue modulo 11, we only consider Case 2.
   - From \( 11^{\frac{p-1}{2}} + 1 = 2pz^2 \), we have \( p \mid 11^{\frac{p-1}{2}} + 1 \). Thus, \( 2y^2 = 11^{\frac{p-1}{2}} - 1 \equiv -2 \pmod{p} \), implying \( p \mid y^2 + 1 \).
   - Therefore, \( 4 \mid p-1 \). Let \( k = \frac{p-1}{4} \). Then \( 11^{\frac{p-1}{4}} + 1 = u^2 \) or \( 11^{\frac{p-1}{4}} + 1 = 2u^2 \).
   - Both cases are impossible because \( u-1 \) and \( u+1 \) cannot both be powers of 11, and 2 is not a quadratic residue modulo 11.

3. **Conclusion for Part (b)**:
   Therefore, there is no prime \( p \) for which \( \frac{11^{p-1} - 1}{p} \) is a perfect square.

The final answer is \( \boxed{ p = 3 } \) for part (a) and no prime \( p \) for part (b).