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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 $(x^2)/2$?

Medhaj Patel
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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.

Aditya Guha Roy
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    $f(z) = (z^2)/2$ is not a solution. In that case, $f(4) = 8$ which is not $10$ – Medhaj Patel Nov 28 '22 at 06:20
  • Yeah I didn't consider the initial condition; my answer was just meant to address the original functional equation. The part "Plug in this solution to see what you get" was essentially meant to verify the initial conditions and all, to see if the solution satisfies them or not. In this case it doesn't but often it gives us an idea of what could be a solution; in other words, solving a functional equation leaving aside the initial conditions is often good to come up with a guess, and then working on the guess to get the solution is often a good strategy. – Aditya Guha Roy Nov 30 '22 at 05:54