The general solution is $f(a) = \frac{a(a+k)}{2}$ for odd integers $k$. It is easy to show by substitution that the stated formula is a solution. We now show it is the only solution.

Putting $a = b = 0$ gives $f(0) = 0$. Putting $b = 1$ gives
$$f(a + 1) = f(a) + f(1) + a.$$
By induction on $a$ it now follows that for all integers $a \ge 0$ we have
$$f(a) = af(1) + \frac{(a-1)a}{2} = \frac{a(a + 2f(1) - 1)}{2}.$$
This is of the claimed form with $k = 2f(1) - 1$ which is of course odd.

Finally, using $b = -a$ in the original functional equation, we get
$$0 = f(0) = -a^2 + f(a) + f(-a),$$
hence for $a > 0$
$$f(-a) = a^2 - f(a) = a^2 - \frac{a(a + k)}{2} = \frac{(-a)(-a + k)}{2}$$
and so the claimed form holds for all integers $a$.