A function $f:\mathbb{Z} \to \mathbb{Z}$ satisfies \begin{align*} f(x+4)-f(x) &= 8x+20, \\ f(x^2-1) &= (f(x)-x)^2+x^2-2 \end{align*}for all integers $x.$ Enter the ordered pair $(f(0),f(1)).$
+287 It turns out that the second constraint alone completely determines the values of $f(-1)$, $f(0)$, and $f(1)$. The first constraint just makes it possible to determine the values of $f$ at other domain points.
We set $x = 0, 1, -1$ in order in the second constraint to get these systems of equations
\begin{eqnarray*}
f(-1) &=& f(0)^2-2 \\
f(0) &=& (f(1)-1)^2+1-2 \\
f(0) &=& (f(-1)+1)^2+1-2
\end{eqnarray*}
Solving we find 3 solutions $(f(-1), f(0), f(1)) = (-1,-1,1)$ or $(-2,0,0)$ or $(-2,0,2)$.
The answer you are looking for is any one of $(-1,1)$, $(0,0)$, or $(0,2)$.
