Constructing an arithmetic progression so that $\sum_{i=1}^n f(x_i) =0$ Let $f: \mathbb{R} \to \mathbb{R} $ be a continuous function so that $ \exists a, b \in \mathbb{R} $ with $f(a) f(b) <0$. Prove that $\forall n>2 \exists$ an arithmetic progression $x_1<x_2<...<x_n$ so that $\sum_{i=1}^n f(x_i) =0$.
What I have observed is that there $\exists c \in (a, b) $ so that $f(c) =0$. I think this is what we need to use to construct that arithmetic progression, but I can't manage to do it. 
 A: Let $n\ge 1$ be given. We may assume w.l.o.g. that $a<b$ and $f(a)<0<f(b)$. By the continuity of $f$, there exists $0<\epsilon<\frac{b-a}2$ such that for all $0< h< \epsilon$, $$
f(a+h)<0<f(b-h).
$$ By choosing $\delta>0$ such that $n\delta <\epsilon$, we have that for all $k=0,1,\ldots, n$,
$$
f(a+k\delta)<0<f(b-k\delta) .
$$ Define
$
g(x) = f(x)+f(x+\delta)+\cdots +f(x+n\delta)
$ for $a\le x\le b-n\delta$. Then $g$ is continuous with $g(a)<0<g(b-n\delta)$. By IVT, there is $x_0\in (a,b-n\delta)$ such that
$$
g(x_0) =  f(x_0)+f(x_0+\delta)+\cdots +f(x_0+n\delta)=0,
$$ as wanted.
A: Note that for fixed $n$ we can parameterize such sums by the step size $s$ and the initial value $x_1$. Then we have a map $F: \mathbb{R}^2 \rightarrow \mathbb{R}$ given by $$F(s,x_1) \rightarrow \sum_{i=1}^{n} f(x_1 + s(i-1)),$$
which is clearly continuous (since $f$ is continous) and has a zero at the point $(0,c),$ where $c$ is the zero of $f$ that you observed necessarily exists.
WLOG assume $a$ and $b$ are the same distance from $c$. Take the circle around $(0,c)$ in $\mathbb{R}^2$ of radius $|(a-c)|$. Then one of $F(0,a)$ and $F(0,b)$ is negative, the other is positive, and the restriction of $F$ to this circle is continuous, and so there must be a zero of $F$ on the circle with $s$ nonzero (since we've already shown that the two points on the circle with $s=0$ are not zeroes of $F$). This implies the existence of your arithmetic progression.
Can this be cleaned up?
