$f(1+x) = f(x)$ for all real $x$, $f$ is a polynomial and $f(5) = 11$. What is $f\large(\frac{15}{2}\large)$?

Question

I was looking through a GRE math subject test practice test (here) and in particular I was confused regarding this question

I chose E because I thought that (since the problem didn't specify) it could be an infinite polynomial that becomes the Taylor expansion of $f(x) = A \sin (x\pi)+11$, where $A$ can be anything.

The correct answer is apparently C. Is there a reason (other than my probably unwarranted assumption) that the answer HAS to be C?

Answer 1

If $f$ isn't a constant polynomial then for every root $x_0$ of it $x_0+n$ is also a root for all $n$ and this contradicts the fundamental theorem of algebra. Conclude.

Answer 2

$$f(x) = a_0 + a_1x^1 + \cdots + a_nx^n$$

$$f(x+1) = a_0 + a_1(x+1)^1 + \cdots + a_n(x+1)^n$$

It can be shown by induction that

$$f(x) = f(x+1) \to a_1=a_2=\cdots=a_n=0$$

Note that

$$f(x) = a_0 + a_1x^1 + \cdots$$

is not a polynomial

Answer 3

It is C. You have the constant map because that is the only polynomial that will give you back f(x)=f(1+x)=f(2+x)... as a function value infinitely many times! Every other polynomial "goes somewhere" on the tail (plus or minus infinity).