If $f(x + y) = f(x) + f(y) +xy$ and $f(4) = 10$ then $f(2001) = ?$ Why is my answer wrong?

Question

I saw this question on YouTube: https://www.youtube.com/watch?v=h98AElJPxa8

$f(x+y) = f(x) + f(y) +xy$, and $f(4) = 10$ then $f(2001) = ?$

The following is the way I solved it, and it looks right to me but they have a different solution in the video:

$f(a) = (a^2)/2 + 2$ => $f(x+y) = (x^2 + 2xy + y^2)/2 + 2$ => $f(x+y) = (x^2)/2 + 2 + (y^2)/2 + xy$

So that we have $f(x+y) = f(x) + f(y) +xy$, I define $f(x) = (x^2)/2 + 2$ and $f(y) = (y^2)/2$

Using: $f(a) = (a^2)/2 + 2$, $f(4) = (4^2)/2 + 2 = 16/2 +2 = 8 + 2 = 10$

So if $a = 2001$ then $f(2001) = 2002002.5$ or $f(2000+1) = 2002002.5$

Which is different from the video. Are there multiple answers? Where am I going wrong? can $f(x)$ have a constant of $+2$ beside $(x^2)/2$?

Answer 1

While finding whether $f(x) = x^2/2 + 2$ satisfies the given functional equation,you accidentally plugged $f(y) = y^2/2$ when it should have been $y^2+2;$ can you now verify whether your solution is valid or not?

Well now, returning to the functional equation $f(x+y) = f(x) + f(y) + xy$ notice that $f(z) = z^2/2$ is a possible solution to this. Plug in this solution to see what you get.