Question 1209928
Let the given equation be
$$f(a+b) = f(a) + f(b) - ab$$
for all nonnegative integers $a$ and $b$, and $f(1) = 7$.

First, we find $f(2)$:
$$f(2) = f(1+1) = f(1) + f(1) - (1)(1) = 7 + 7 - 1 = 13$$
Next, we find $f(3)$:
$$f(3) = f(2+1) = f(2) + f(1) - (2)(1) = 13 + 7 - 2 = 18$$
Next, we find $f(4)$:
$$f(4) = f(3+1) = f(3) + f(1) - (3)(1) = 18 + 7 - 3 = 22$$

Let's try to find a pattern.
$f(1) = 7 = 1^2 + 6(1)$
$f(2) = 13 = 2^2 + 6(2) + 1$
$f(3) = 18 = 3^2 + 6(3) - 9 = 9+18 = 27$
$f(3) = 18 = 3^2 + 6(3) = 9 + 9 = 18$
$f(4) = 22 = 4^2 + 6(4) - 18 = 16+24-18=22$

We can guess that $f(x) = x^2 + 6x$.
Let's check if this satisfies the given equation.
$$f(a+b) = (a+b)^2 + 6(a+b) = a^2 + 2ab + b^2 + 6a + 6b$$
$$f(a) + f(b) - ab = (a^2 + 6a) + (b^2 + 6b) - ab = a^2 + 6a + b^2 + 6b - ab$$
These are not equal. Let's try to find a pattern.

Let $f(x) = Ax^2 + Bx$.
$f(1) = A+B = 7$
$f(2) = 4A + 2B = 13$
Multiply the first equation by 2:
$2A + 2B = 14$
Subtract this from the second equation:
$2A = -1$
$A = -1/2$
$B = 7 - A = 7 + 1/2 = 15/2$
Thus $f(x) = -\frac{1}{2} x^2 + \frac{15}{2} x$.

Check:
$f(a+b) = -\frac{1}{2} (a+b)^2 + \frac{15}{2} (a+b) = -\frac{1}{2} (a^2 + 2ab + b^2) + \frac{15}{2} a + \frac{15}{2} b$
$f(a) + f(b) - ab = (-\frac{1}{2} a^2 + \frac{15}{2} a) + (-\frac{1}{2} b^2 + \frac{15}{2} b) - ab = -\frac{1}{2} a^2 - \frac{1}{2} b^2 + \frac{15}{2} a + \frac{15}{2} b - ab$
$f(a+b) = f(a) + f(b) - ab$
So $f(x) = -\frac{1}{2} x^2 + \frac{15}{2} x$.

Now, compute $f(123)$:
$f(123) = -\frac{1}{2} (123)^2 + \frac{15}{2} (123) = \frac{1}{2} (15(123) - 123^2) = \frac{123}{2} (15 - 123) = \frac{123}{2} (-108) = 123(-54) = -6642$

Final Answer: The final answer is $\boxed{-6642}$