Question 1209952
We are given the functional equation:

\[
f(x) + f(y) = f(x + y) - 4xy - 1 + f(x^2) + f(y^2)
\]

for all integers \( x, y \), along with the initial condition \( f(1) = 1 \). Our goal is to determine \( f(n) \) explicitly.

### Step 1: Plugging in \( x = 0, y = 0 \)

Substituting \( x = 0, y = 0 \) into the given equation:

\[
f(0) + f(0) = f(0) - 4(0)(0) - 1 + f(0^2) + f(0^2)
\]

Since \( f(0^2) = f(0) \), we simplify:

\[
2f(0) = f(0) - 1 + 2f(0)
\]

Canceling \( 2f(0) \) from both sides, we get:

\[
0 = -1 \Rightarrow f(0) = 0.
\]

### Step 2: Plugging in \( y = 1 \)

Setting \( y = 1 \) in the functional equation:

\[
f(x) + f(1) = f(x + 1) - 4x(1) - 1 + f(x^2) + f(1^2).
\]

Since \( f(1) = 1 \) and \( f(1^2) = f(1) = 1 \), we substitute:

\[
f(x) + 1 = f(x + 1) - 4x - 1 + f(x^2) + 1.
\]

Simplify:

\[
f(x) = f(x + 1) - 4x + f(x^2).
\]

### Step 3: Plugging in \( x = 1, y = 1 \)

Setting \( x = y = 1 \):

\[
f(1) + f(1) = f(2) - 4(1)(1) - 1 + f(1^2) + f(1^2).
\]

Using \( f(1) = 1 \):

\[
1 + 1 = f(2) - 4 - 1 + 1 + 1.
\]

\[
2 = f(2) - 3.
\]

\[
f(2) = 5.
\]

### Step 4: Plugging in \( x = 2, y = 1 \)

\[
f(2) + f(1) = f(3) - 4(2)(1) - 1 + f(2^2) + f(1^2).
\]

Using \( f(2) = 5 \), \( f(1) = 1 \), \( f(1^2) = 1 \):

\[
5 + 1 = f(3) - 8 - 1 + f(4) + 1.
\]

\[
6 = f(3) - 8 + f(4) + 1.
\]

\[
5 = f(3) - 8 + f(4).
\]

Rearrange:

\[
f(3) + f(4) = 13.
\]

### Step 5: Identifying a Pattern

Based on observed values \( f(1) = 1 \), \( f(2) = 5 \), and the recurrence relation, we suspect \( f(n) \) follows a quadratic pattern. Assume:

\[
f(n) = an^2 + bn + c.
\]

Using \( f(1) = 1 \):

\[
a(1)^2 + b(1) + c = 1 \Rightarrow a + b + c = 1.
\]

Using \( f(2) = 5 \):

\[
a(2)^2 + b(2) + c = 5 \Rightarrow 4a + 2b + c = 5.
\]

Using \( f(0) = 0 \):

\[
a(0)^2 + b(0) + c = 0 \Rightarrow c = 0.
\]

So,

\[
a + b = 1, \quad 4a + 2b = 5.
\]

Solving for \( a, b \):

\[
b = 1 - a,
\]

\[
4a + 2(1 - a) = 5.
\]

\[
4a + 2 - 2a = 5.
\]

\[
2a = 3.
\]

\[
a = \frac{3}{2}, \quad b = 1 - \frac{3}{2} = -\frac{1}{2}.
\]

Thus, 

\[
f(n) = \frac{3}{2} n^2 - \frac{1}{2} n.
\]

Verifying,

\[
f(n) = \frac{3}{2} n(n - 1).
\]

Final Answer:

\[
\boxed{f(n) = \frac{3}{2} n(n-1).}
\]