The pattern ends here

let $p(n)=f(n)-17n$ we see that 1,2,3 are roots of p(n). let a be the other root, since p is monic: $p(n)=(n-1)(n-2)(n-3)(n-a)\Longrightarrow f(n)=(n-1)(n-2)(n-3)(n-a)+17n$ put this in the given expression: $f(0)+f(4)=(-1)(-2)(-3)(-a)+17*0+(3)(2)(1)(4-a)+17*4$ $=6a+24-6a+68=\boxed{92}$

I looked at the first and second differences. Nothing is known about the third differences, but because the polynomial is fourth order the fourth differences are all equal, and because it is monic they are all 24:

$\begin{array}{ccccccccc} ?? & & 17 & & 34 & & 51 & & ?? \\ & ?? & & 17 & & 17 & & ?? & \\ & & ?? & & 0 & & ?? & & \\ & & & ?? & & ?? & & & \\ \cdots & & (24) & & 24 & & (24) & & \cdots \end{array}$

So we cannot determine $f(0)$ and $f(4)$ separately--there are infinitely many solutions--but hopefully we can uniquely determine their sum. Call the first missing third difference $x$ , then

$\begin{array}{ccccccccc} ?? & & 17 & & 34 & & 51 & & ?? \\ & ?? & & 17 & & 17 & & ?? & \\ & & ?? & & 0 & & ?? & & \\ & & & x & & 24+x & & & \\ & & & & 24 & & & & \end{array}$

and working our way up we get

$\begin{array}{ccccccccc} -x & & 17 & & 34 & & 51 & & 92+x \\ & 17+x & & 17 & & 17 & & 41+x & \\ & & -x & & 0 & & 24+x & & \\ & & & x & & 24+x & & & \\ & & & & 24 & & & & \end{array}$

Reading off we have therefore

$f(0) + f(4) = (-x) + (92 + x) = \boxed{92}.$

We can write $f(x)=x^4+P_3(x)$ . Using the data that we got, we can get the following system of equations: $P_3(1)=16, P_3(2)=18, P_3(3)=-30$

Now we can set $P_3(x)=ax^3+bx^2+cx$ or any other of the four combinations of coefficients and powers of x. This particular one gives us $a=-6, b=11=c$ . This works, and combining it with $x^4$ we get the correct answer.