For a sequence $ a_i \in \mathbb{R}, i \in \{0, 1, 2, \ldots\}$ we have $ a_0 \equal{} 1$ and  \[ a_{n\plus{}1} \equal{} a_n \plus{} \sqrt{a_{n\plus{}1} \plus{} a_n} \quad \forall n \in \mathbb{N}.\] Prove that this sequence is unique and find an explicit formula for this recursively defined sequence.
1. **Initial Condition and Recurrence Relation:**
   We start with the initial condition \( a_0 = 1 \) and the recurrence relation:
   \[
   a_{n+1} = a_n + \sqrt{a_{n+1} + a_n} \quad \forall n \in \mathbb{N}.
   \]

2. **Isolating \( a_{n+1} \):**
   To isolate \( a_{n+1} \), we let \( x = a_{n+1} \). Then the recurrence relation becomes:
   \[
   x = a_n + \sqrt{x + a_n}.
   \]
   Squaring both sides to eliminate the square root, we get:
   \[
   x^2 = (a_n + \sqrt{x + a_n})^2.
   \]
   Expanding the right-hand side:
   \[
   x^2 = a_n^2 + 2a_n\sqrt{x + a_n} + (x + a_n).
   \]
   Simplifying, we have:
   \[
   x^2 = a_n^2 + x + 3a_n + 2a_n\sqrt{x + a_n}.
   \]
   Rearranging terms:
   \[
   x^2 - x - a_n^2 - 3a_n = 2a_n\sqrt{x + a_n}.
   \]
   Squaring both sides again to eliminate the square root:
   \[
   (x^2 - x - a_n^2 - 3a_n)^2 = 4a_n^2(x + a_n).
   \]

3. **Finding the Explicit Formula:**
   We hypothesize that the sequence \( a_n \) can be expressed in a closed form. Let's assume:
   \[
   a_n = \frac{n(n + 3)}{2} + 1.
   \]
   We will verify this hypothesis by induction.

4. **Base Case:**
   For \( n = 0 \):
   \[
   a_0 = \frac{0(0 + 3)}{2} + 1 = 1,
   \]
   which matches the initial condition.

5. **Inductive Step:**
   Assume \( a_n = \frac{n(n + 3)}{2} + 1 \) holds for some \( n \). We need to show it holds for \( n + 1 \):
   \[
   a_{n+1} = a_n + \sqrt{a_{n+1} + a_n}.
   \]
   Using the hypothesis:
   \[
   a_n = \frac{n(n + 3)}{2} + 1.
   \]
   We need to find \( a_{n+1} \):
   \[
   a_{n+1} = \frac{n(n + 3)}{2} + 1 + \sqrt{a_{n+1} + \frac{n(n + 3)}{2} + 1}.
   \]
   Let \( a_{n+1} = \frac{(n+1)((n+1) + 3)}{2} + 1 \):
   \[
   a_{n+1} = \frac{(n+1)(n + 4)}{2} + 1 = \frac{n^2 + 5n + 4}{2} + 1 = \frac{n^2 + 5n + 6}{2}.
   \]
   We need to verify:
   \[
   \frac{n^2 + 5n + 6}{2} = \frac{n(n + 3)}{2} + 1 + \sqrt{\frac{n^2 + 5n + 6}{2} + \frac{n(n + 3)}{2} + 1}.
   \]
   Simplifying the right-hand side:
   \[
   \frac{n(n + 3)}{2} + 1 + \sqrt{\frac{n^2 + 5n + 6 + n^2 + 3n + 2}{2}} = \frac{n(n + 3)}{2} + 1 + \sqrt{\frac{2n^2 + 8n + 8}{2}} = \frac{n(n + 3)}{2} + 1 + \sqrt{n^2 + 4n + 4} = \frac{n(n + 3)}{2} + 1 + (n + 2).
   \]
   Simplifying further:
   \[
   \frac{n(n + 3)}{2} + 1 + n + 2 = \frac{n^2 + 3n}{2} + n + 3 = \frac{n^2 + 5n + 6}{2}.
   \]
   This confirms our hypothesis.

6. **Conclusion:**
   The sequence \( a_n \) is unique and given by:
   \[
   a_n = \frac{n(n + 3)}{2} + 1.
   \]

The final answer is \( \boxed{ a_n = \frac{n(n + 3)}{2} + 1 } \).