For the function $f$, we know that for all integers $x, y$,

$$
f\left(x-y^{2}\right)=f(x)+\left(y^{2}-2 x\right) \cdot f(y)
$$

Determine the value of $f(1984)$.
For the function $f$, we know that for every integer $x$, $y$,

$$
f\left(x-y^{2}\right)=f(x)+\left(y^{2}-2 x\right) \cdot f(y)
$$

Determine the value of $f(1984)$.

Solution. If $y=1$, then according to (1), for every integer $x$,

$$
f(x-1)=f(x)+(1-2 x) \cdot f(1)
$$

Let's substitute the positive integers from 1 to $n$ for $x$ in (2):

$$
\begin{gathered}
f(0)=f(1)-f(1) \\
f(1)=f(2)-3 \cdot f(1) \\
\vdots \\
f(n-1)=f(n)-(2 n-1) \cdot f(1)
\end{gathered}
$$

Adding these equations, the values $f(1), f(2), \ldots, f(n-1)$ appear on both sides, so after rearranging, we get

$$
f(0)=f(n)-S_{n} \cdot f(1)
$$

where $S_{n}$ is the sum of the first $n$ positive odd numbers. It is known that $S_{n}=n^{2}$, and according to (3), $f(0)=0$, so if $n$ is a positive integer, then

$$
f(n)=f(1) \cdot n^{2}
$$

However, we cannot say more than this. Let $a$ be any real number, and consider the function $f(x)=a \cdot x^{2}$. Here, $f(1)=a$. On the other hand,

$$
f(x)+\left(y^{2}-2 x\right) \cdot f(y)=a x^{2}+\left(y^{2}-2 x\right) a y^{2}=a\left(x^{2}+y^{4}-2 x y^{2}\right)=a\left(x-y^{2}\right)^{2}
$$

which is nothing but $f\left(x-y^{2}\right)$. Thus, if $f(x)=a x^{2}$, then (1) holds for $f$, and therefore nothing can be concluded about $f(1)$ from the fact that (1) holds for $f$.

Therefore, the value of $f(1984)$ can be arbitrary, since for every real number $A$, there is a function $f$ that satisfies (1) and $f(1984)=A$, namely $f(x)=\frac{A}{1984^{2}} x^{2}$.