If a function from $\mathbb{R}$ to $\mathbb{R}$ is periodic with any $P\in\mathbb{R}$, is it necessarily constant?

Question

Suppose I've been given a function $f:\mathbb{R}\rightarrow\mathbb{R}$ with the property that for all $x,P\in\mathbb{R}$, $f(x+P)=f(x)$. Does it follow that $f$ is a constant function?

I've been trying to come up with a counterexample to this claim for a few minutes, but have had no success. Intuitively, it seems like the answer should be yes because fixing an arbitrary $x$ and letting $P$ vary in $\mathbb{R}$ should make the quantity $x+P$ sweep through all possible real numbers, so then the statement $f(x+P)=f(x)$ would boil down to $f(x_0)=f(x)$ (here, $x_0$ is any real number), which I think is only satisfied when $f$ is constant.

I'd appreciate a rigorous proof of the truth or falsity of this claim.

Answer 1

$$f(y)=f(x+(y-x))=f(x)$$ So with $P=y-x$ we obtain thę thesis.

Answer 2

Yes it is constant. $$f(x)=f(0+x)=f(0)$$ for any $x\in\mathbb R$, using $P=x$.