Find all solution $(p,r)$ of the "Pythagorean-Euler Theorem" $$p^p+(p+1)^p+\cdots+(p+r)^p=(p+r+1)^p$$Where $p$ is a prime and $r$ is a positive integer.

[i]Proposed by Li4 and Untro368[/i]
1. **Case \( p = 2 \):**
   \[
   2^2 + (2+1)^2 + \cdots + (2+r)^2 = (3+r)^2
   \]
   Simplifying, we get:
   \[
   \sum_{k=2}^{2+r} k^2 = (3+r)^2
   \]
   Using the formula for the sum of squares:
   \[
   \sum_{k=2}^{2+r} k^2 = \frac{(2+r)(3+r)(2r+5)}{6} - 1
   \]
   Setting this equal to \((3+r)^2\):
   \[
   \frac{(2+r)(3+r)(2r+5)}{6} - 1 = (3+r)^2
   \]
   This implies:
   \[
   r+3 \mid 6 \Rightarrow r = 3
   \]
   However, checking:
   \[
   2^2 + 3^2 + 4^2 + 5^2 \neq 6^2
   \]
   Thus, no solutions for \( p = 2 \).

2. **Case \( p \geq 3 \):**
   Let \( q \) be an odd prime divisor of \( r+1 \). Then \( r+1 = lq \) for some positive integer \( l \). Since \( p, p+1, \ldots, p+r \) are \( r+1 \) consecutive integers:
   \[
   p^p + (p+1)^p + \cdots + (p+r)^p \equiv l(1^p + 2^p + \cdots + (q-1)^p) \pmod{q}
   \]
   Let \( g \) be a primitive root modulo \( q \). Then:
   \[
   1^p + 2^p + \cdots + (q-1)^p \equiv g^p + (g^p)^2 + \cdots + (g^p)^{q-1} \pmod{q}
   \]
   If \( g^p \equiv 1 \pmod{q} \), then \( q-1 \mid p \), which is impossible since \( p \) is odd. Hence:
   \[
   g^p \not\equiv 1 \pmod{q}
   \]
   Therefore:
   \[
   g^p + (g^p)^2 + \cdots + (g^p)^{q-1} \equiv \frac{g^p((g^p)^{q-1} - 1)}{g^p - 1} \equiv 0 \pmod{q}
   \]
   So:
   \[
   q \mid p^p + (p+1)^p + \cdots + (p+r)^p = (p+r+1)^p \equiv p^p \pmod{q}
   \]
   Thus, \( q = p \).

3. **Considering \( r+1 = 2^b p^a \) for some \( a, b \geq 0 \):**
   If \( 4 \mid r+1 \), then:
   \[
   1 \equiv (p+r+1)^p = p^p + (p+1)^p + \cdots + (p+r)^p \equiv p + (p+1) + \cdots + (p+r) \equiv (r+1)p + \frac{r(r+1)}{2} \equiv 0 \pmod{2}
   \]
   This is a contradiction. So \( r+1 = p^a \) or \( r+1 = 2p^a \).

4. **Case \( r+1 = 2p^a \):**
   If \( a = 0 \):
   \[
   p^p + (p+1)^p = (p+2)^p \Rightarrow 1 = 2^p \pmod{p}
   \]
   This is a contradiction. So \( a \geq 1 \).

5. **Let \( q \) be a prime divisor of \( r \), \( q \neq 2 \):**
   \[
   p^p + (p+1)^p + \cdots + (p+r)^p \equiv l(1^p + \cdots + (q-1)^p) + p^p \pmod{q}
   \]
   As before, \( q \mid 1^p + \cdots + (q-1)^p \), so:
   \[
   p^p \equiv (p+1)^p \pmod{q} \Rightarrow q \mid z^p - 1 \text{ where } z \equiv p(p+1)^{-1} \pmod{q}
   \]
   Now \( q \mid z^p - 1 \Rightarrow \text{ord}_q z \mid p \), so \( \text{ord}_q z = 1 \) or \( \text{ord}_q z = p \). If \( \text{ord}_q z = 1 \):
   \[
   p = p+1 \pmod{q}
   \]
   This is impossible. So \( p = \text{ord}_q z \mid q-1 \).

6. **If \( r = 2p^a - 1 = q_1^{a_1} q_2^{a_2} \cdots q_t^{a_t} \):**
   \[
   q_i \mid r \Rightarrow q_i = 1 \pmod{p} \text{ for } i = 1, 2, \ldots, t
   \]
   This implies:
   \[
   -1 = 2p^a - 1 = 1 \cdot 1 \cdots 1 = 1 \pmod{p} \Rightarrow p \mid 2
   \]
   This is a contradiction.

7. **Case \( r+1 = p^a \):**
   Using the same argument as before, we get:
   \[
   r = p^a - 1 = 2^m q_1^{a_1} q_2^{a_2} \cdots q_t^{a_t} \equiv 2^m \pmod{p}
   \]
   So \( p \mid 2^m + 1 \) and \( 2^m \mid p^a - 1 \).

8. **Let \( q \) be an odd prime factor of \( r+2 \):**
   \[
   r+2 = ql
   \]
   Then:
   \[
   (p-1)^p \equiv (p+r+1)^p = p^p + (p+1)^p + \cdots + (p+r)^p \equiv l(1^p + \cdots + (q-1)^p) - (p+r+1)^p \equiv -(p-1)^p \pmod{q}
   \]
   So \( q \mid p-1 \), but also \( q \mid r+2 = p^a + 1 \), thus \( q \mid 2 \). This is a contradiction.

9. **Conclusion:**
   \[
   r+2 = 2^t \text{ for some } t \geq 2
   \]
   \[
   2^m \mid p^a - 1 = 2^t - 2 \Rightarrow m = 1
   \]
   So:
   \[
   p \mid 2^m + 1 = 3 \Rightarrow p = 3
   \]
   Now:
   \[
   3^a + 1 = 2^t
   \]
   Since \( 4 \mid 3^a + 1 \), \( a \) is odd, and thus:
   \[
   3^a + 1 = 4 \pmod{8} \Rightarrow t = 2
   \]
   Therefore:
   \[
   r = 2
   \]
   Indeed:
   \[
   3^3 + 4^3 + 5^3 = 6^3
   \]
   So \( (p, r) = (3, 2) \) is the only solution.

The final answer is \( \boxed{ (p, r) = (3, 2) } \)