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I'm asked to answer this question:

Let $f$ be a polynomial function not equal to zero, what can be said about $f(x)$ having $f(x-1)=f(x)$

Which has four choices, $f(x)$ is an odd or even or negative or positive function.

What I've done is finding out that $f(1)=f(0)$ and I'm stuck here.

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Let $f(0)=c$. By what you did, $f(1)=c$. By putting $x=2$, $f(2)=c$. By putting $x=3$, $f(3)=c$. And so on. But a polynomial of degree $n$ is completely determined by its values at $n+1$ points. So $f(x)=c$ for all $x$. Thus $f$ is even. It could be also odd (if $c=0$), or negative (if $c$ is, or positive (if $c$ is), but does not need to be. – André Nicolas Jan 23 '12 at 12:11
@AndréNicolas: Thanks, that solves the problem as a complete answer. – Gigili Jan 23 '12 at 12:14
up vote 5 down vote accepted

HINT: If $c$ is a zero of $f$, then so is $c-1$, $c-2$, ...

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Do you mean $f(x)=0$? – Gigili Jan 23 '12 at 8:42
Or if you're uncomfortable assuming that $f$ has a zero, consider the polynomial $f(x)-f(0)$. – Jonas Meyer Jan 23 '12 at 8:43
@zizi A number $c$ is said to be a zero of a polynomial $f$ if $f(c) = 0$. In other words, $c$ is a root/solution of the polynomial equation $f(x)=0$. – Srivatsan Jan 23 '12 at 8:45
@Srivatsan I see, I misinterpreted the answer, sorry. It means that $\forall x: f(x)=c \neq 0$. Am I right? so all of the above can be true about $f$ except being odd. – Gigili Jan 23 '12 at 8:51
@zizi You are using $c$ in a different sense. Dan's hint implies that $f(x)$ is a constant $C$ independent of $x$; moreover the question says that this constant $C$ is nonzero. Thus $f(x)$ is certainly always even and never odd. As to the other two options, depending on $C$, $f(x)$ could be always positive or always negative; this cannot be deduced unless we are told the sign of $C$. So, if exactly one option is correct, then that ought to be "even". [Note: I am using $C$ to denote the constant, since $c$ is used by Dan to denote the zeroes.] – Srivatsan Jan 23 '12 at 8:57

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