We are given the functional equation:

$f(x + y) = f(x) + f(y) + 2xy,$

along with the initial condition $f(1) = 1$. We are tasked with finding $f(25)$.

Step 1: Analyze the functional equation

Set $x = y = 0$ in the equation: $f(0 + 0) = f(0) + f(0) + 2(0)(0).$ This simplifies to: $f(0) = 2f(0).$ Thus, $f(0) = 0$.

Step 2: Substitute $y = 1$

Substitute $y = 1$ in the functional equation: $f(x + 1) = f(x) + f(1) + 2x(1).$ Using $f(1) = 1$, this becomes: $f(x + 1) = f(x) + 1 + 2x.$

Step 3: Recursively compute $f(x)$

From the above recurrence relation, we compute $f(2)$: $f(2) = f(1 + 1) = f(1) + 1 + 2(1) = 1 + 1 + 2 = 4.$

Similarly, compute $f(3)$: $f(3) = f(2 + 1) = f(2) + 1 + 2(2) = 4 + 1 + 4 = 9.$

We observe a pattern that $f(n) = n^2$. Let us verify this general formula.

Step 4: Verify the general formula $f(n) = n^2$

Assume $f(x) = x^2$. Substitute into the functional equation: $f(x + y) = (x + y)^2, \quad f(x) + f(y) + 2xy = x^2 + y^2 + 2xy.$ Both sides are equal, so $f(x) = x^2$ satisfies the functional equation.

Step 5: Compute $f(25)$

Using $f(x) = x^2$, we find: $f(25) = 25^2 = 625.$

Final Answer:

$\boxed{625}$

Would you like further details on this? Here are 5 questions related to this topic:

1. How can functional equations be solved systematically?
2. What are some common patterns in functional equations?
3. Can the verification step differ for more complex functional equations?
4. How can initial conditions affect the solution to functional equations?
5. Are there other examples where $f(x) = x^2$ satisfies a functional equation?

Tip: When solving functional equations, try substituting specific values like $x = 0$ or $y = 1$ to simplify and find key properties of the function.