What is the value of $f(0)+f(8)$? Suppose $f$ is a polynomial of degree $7$ which satisfies $f(1) =2$, $f(2)=5$, $f(3)=10$, $f(4)=17$, $f(5)=26$, $f(6)=37$ and $f(7)=50$. What is the value of $f(0)+f(8)$?
 A: The given values are the values of the quadratic polynomial $X^2+1$. Therefore, we have $f(X) - X^2 - 1$ a polynomial of degree $7$ with the seven zeros $1,2,3,4,5,6,7$, so
$$f(X) = c\cdot\prod_{k=1}^7 (X-k) + X^2+1.$$
with an unknown $c$ ($c\neq 0$ if the degree is exactly $7$). Then we have
$$f(8) = c\cdot 7! + 65$$
and
$$f(0) = c\cdot(-1)^7 7! + 1,$$
whence $$f(0) + f(8) = 65 + 1 + c(7! - 7!) = 66.$$
A: Note that the differences between the given values form an arithmetic sequence:
$$ +3,\ +5,\ +7,\ +9,\ +11,\ +13\,.$$
So that, we have an easy solution $f$, having $f(0)=1$ and $f(8)=65$. 
(Anyway, $f(x)=x^2+1$.)
Note that $7$ values determine uniquely a polynomial of degree $\le \bf 6$.
So, any other polynomial that satisfies the conditions is of the form
$$f(x)= x^2-1\ + \ g(x)\cdot(x-1)(x-2)\dots(x-7)\,.$$
A: As stated, this was a problem posed on Brilliant. The following is the solution written up by the member on Brilliant who posed this problem.

A: Note that the polynomial $f$ will not be uniquely defined, but $f(0)+f(8)$ will be. Indeed, given two possible polynomials, how do they differ?
A: There's a straightforward approach one can use to recover a polynomial from its differences. Let's use a polynomial of degree at most 3 to demonstrate.
Such a polynomial is determined by 4 values. Let's say we have the consecutive values


*

*$f(0) = 1$

*$f(1) = 2$

*$f(2) = 4$

*$f(3) = 9$


Then we can make a diagram of the differences
$$
\begin{matrix}
\\ & & & 0
\\ & & 2 & & 2
\\ & 1 & & 3 & & 5
\\ 1 & & 2 & & 4 & & 9
\end{matrix}
$$
Because we know the polynomial is of degree at most $3$, that means if we extend this diagram to the left and right, the top row will be constant: in this case it will be $0$ everywhere.
This actually means the second row will be constant as well -- this means our polynomial is actually quadratic.
We could work to recover $f$ -- e.g. with a Newton series. But if we want $f(4)$, there's a simple method: just compute the next diagonal in the picture:
$$
\begin{matrix}
\\ & & & 0 & & 0
\\ & & 2 & & 2 & & 2
\\ & 1 & & 3 & & 5 & & 7
\\ 1 & & 2 & & 4 & & 9 & & 16
\end{matrix}
$$
and so $f(4) = 16$.
The original problem should be doable with a similar technique. However, you must do something appropriate to account for the fact you only have an explicit decimal value for $7$ terms, rather than $8$!
