Find all positive integers $a, b, c, d,$ and $n$ satisfying $n^a + n^b + n^c = n^d$ and prove that these are the only such solutions.
We are given the equation \( n^a + n^b + n^c = n^d \) and need to find all positive integers \( a, b, c, d, \) and \( n \) that satisfy it. We will also prove that these are the only solutions.

1. **Assume \( n > 1 \) and \( a \leq b \leq c \):**
   Without loss of generality, we can assume \( a \leq b \leq c \). We can write \( b = a + x \) and \( c = b + y = a + x + y \) where \( x, y \in \mathbb{N}_0 \).

2. **Rewrite the equation:**
   \[
   n^a + n^b + n^c = n^d \implies n^a + n^{a+x} + n^{a+x+y} = n^d
   \]
   Factor out \( n^a \) from the left-hand side:
   \[
   n^a (1 + n^x + n^{x+y}) = n^d
   \]
   This implies:
   \[
   n^d = n^a (1 + n^x + n^{x+y})
   \]

3. **Consider the case \( x \in \mathbb{N} \):**
   If \( x \in \mathbb{N} \), then \( \gcd(1 + n^x + n^{x+y}, n) = \gcd(1, n) = 1 \). This is a contradiction because \( 1 + n^x + n^{x+y} \mid n \) and is greater than 1. Hence, \( x = 0 \).

4. **Simplify with \( x = 0 \):**
   \[
   n^d = n^a (2 + n^y)
   \]
   Now, consider \( y \in \mathbb{N} \).

5. **Case \( y \in \mathbb{N} \):**
   - If \( n \) is odd, then \( \gcd(2 + n^y, n) = \gcd(2, n) = 1 \). This is a contradiction because \( 2 + n^y \mid n^d \) and is greater than 1.
   - Hence, \( n \) must be even. Let \( n = 2k \) where \( k \in \mathbb{N} \).

6. **Substitute \( n = 2k \):**
   \[
   (2k)^d = (2k)^a (2 + (2k)^y)
   \]
   Simplify:
   \[
   2^d k^d = 2^a k^a (2 + 2^y k^y) = 2^a k^a (2 + 2^y k^y)
   \]
   \[
   2^d k^d = 2^a k^a (2 + 2^y k^y)
   \]
   \[
   2^d k^d = 2^{a+1} k^a (1 + 2^{y-1} k^y)
   \]

7. **Consider \( y = 1 \):**
   \[
   2^d k^d = 2^{a+1} k^a (1 + k)
   \]
   Since \( \gcd(1 + k, k) = \gcd(1, k) = 1 \), \( 1 + k \mid 2 \). This implies \( k = 1 \), so:
   \[
   2^d = 2^{a+2}
   \]
   Hence, \( d = a + 2 \). This gives the solution:
   \[
   (a, b, c, d, n) = (a, a, a+1, a+2, 2)
   \]

8. **Consider \( y > 1 \):**
   \[
   1 + 2^{y-1} k^y
   \]
   This term is relatively prime with 2 and \( k \), which is a contradiction because it divides \( 2^d k^d \).

9. **Consider \( y = 0 \):**
   \[
   n^d = 3n^a
   \]
   Let \( k \) be the exponent of 3 in the prime factorization of \( n \). Then \( kd \) is the exponent of 3 in the prime factorization of \( n^d \) and \( 1 + ka \) is the exponent of 3 in the prime factorization of \( 3n^a \). Hence:
   \[
   kd = 1 + ka
   \]
   This implies:
   \[
   k(d - a) = 1
   \]
   So \( k \mid 1 \), hence \( k = 1 \). Therefore, \( d - a = 1 \), so \( d = a + 1 \). This implies:
   \[
   n^{a+1} = 3n^a \implies n = 3
   \]
   Hence, the solution is:
   \[
   (a, b, c, d, n) = (a, a, a, a+1, 3)
   \]

The final answer is \( \boxed{ (a, b, c, d, n) = (a, a, a+1, a+2, 2) } \) and \((a, b, c, d, n) = (a, a, a, a+1, 3)\).