all for a constant , which clearly work.

Obviously the solutions work. Let denote the functional equation. Take and to get

... (1)
... (2)

Substituting both equations into the original becomes



which, by substituting , , is equivalent to the Cauchy functional equation

... (*).

In particular, so .

Next, setting in (1), (2) and equating the two equations, we get

for ... (3)

But for we already know both sides are 0, so (3) still holds.

Now, for any real , put in (3) to evaluate , and then split into more terms with (*):

\[ \begin{align*} f((y+1)^2) &= (y+1)f(y+1) \\ &= (y+1)(f(y)+f(1)) \\ &= yf(y) + f(y) + f(1)y + f(1) \]

For a different evaluation, expand , split into more terms with (*), and then apply (3):

\[ \begin{align*} f((y+1)^2) &= f(y^2 + 2y + 1) \\ &= f(y^2) + 2f(y) + f(1) \\ &= yf(y) + f(y) + f(y) + f(1) \]

And equating the last two equations above results in for all real , so we're done.