Why does $f(x)=f(x+1)$ imply $f(x) \equiv$ constant if $f$ is polynomial? A polynomial function $P: \mathbb{R} \mapsto \mathbb{R}$ has the following property:
$$\forall x\in\mathbb{R} \space \bigg(P(x)=P(x+1)\bigg)$$
Why does this imply that $P \equiv \text{constant}$?
EDIT: When posting the question for the first time I forgot to mention that we are talking about a polynomial function.
 A: Note that the OP has clarified that they are assuming $f$ is a polynomial.
This follows from the standard result that every nonconstant polynomial has only finitely many zeroes: just consider the polynomial $Q(x)=P(x)-P(0)$. We'll have $Q(z)=0$ for every integer $z$, so $Q$ - and hence $P$ itself - must be constant.
Incidentally, the standard result above may feel obvious at first but is surprisingly delicate; you should go through the proof in careful detail.

Alternatively, it's easy to show that if $P$ is a nonconstant polynomial then $\lim_{x\rightarrow\infty}P(x)$ is either $+\infty$ or $-\infty$ depending on the sign of the leading coefficient of $P$. In particular, no nonconstant polynomial is bounded. But every continuous periodic function is bounded, as a consequence e.g. of the Extreme Value Theorem. (This argument was also mentioned by lulu in a comment above, as I was typing this.)
(It's easy to miss the application of EVT here, so let me unpack it. Periodicity gives us that $ran(P)=\{P(x): x\in [0,p]\}$ where $p$ is the period of $P$. Now we need to show that $\{P(x): x\in [0,p]\}$ is bounded. This is where continuity comes in: since $P$ is continuous and $[0,p]$ is a closed and bounded interval, $P$ has a maximum and minimum value on $[0,p]$. Note that we can have discontinuous periodic functions which are unbounded: consider e.g. the extension $\alpha$ of $\tan$ gotten by setting $\alpha(x)=17$ whenever $\tan(x)$ is undefined.)
A: Suppose $P$ has degree $D\ge1$
Differentiate $D-1$ times.
$$P(x)=P(x+1) \implies \frac{d^{D-1} P(x)}{dx^{D-1}}=\frac{d^{D-1} P(x+1)}{dx^{D-1}} $$
Both sides are linear polynomials. Let $P^{(D-1)}(x) = ax+b $. Then, $$ax+b = a(x+1)+ b \implies a=0$$
That’s a contradiction. This means D must be zero.
A: No.
For example,
$$
f(x)=\sin(2\pi x). 
$$
A: It doesn't. Every function with period $1$ has this property.
For example, if $f(x) = \sin 2 \pi x$, then $f(x+1) = \sin (2\pi x +2\pi) = \sin 2\pi x = f(x)$ for all $x$.
A: Suppose there exists nonconstant polynomial $P$. Then it has at least one root and since it is periodic it has infinitely roots. But such polynomial is constant polynomial $P(x)=0$ for all $x$. A contradiction.
A: Let $ n\in\mathbb{N} $, we have for any $ k\in\mathbb{N} $ :$$ P\left(k+1\right)-P\left(k\right)=0\Longrightarrow\sum_{k=0}^{n-1}{\left(P\left(k+1\right)-P\left(k\right)\right)}=0\iff P\left(n\right)=P\left(0\right) $$
Since $ n $ is arbitrary, $ \left(\forall n\in\mathbb{N}\right),\ P\left(n\right)=P\left(0\right) $.
Consider a polynomial function $ Q:x\mapsto P\left(x\right)-P\left(0\right) $.
$ Q $ has an infinite number of zeros, In fact $ \mathbb{N}\subset Q^{-1}{\left(\left\lbrace 0\right\rbrace\right)} $.
Thus : $$ Q=0 $$
Which means : $$ \left(\forall x\in\mathbb{R}\right),\ P\left(x\right)=P\left(0\right) $$
