Fix $k \in \mathbb{N}$, $k \geq 1$. Let $p \in [0,1]$ and $x = (x_0, \ldots, x_k)$ be a $(k+1)$-dimensional *real* vector, and define
$$S(p,x) = -x_0^2 + \sum_{i=0}^k {k \choose i} p^i (1 - p)^{k - i} \cdot (x_i - p)^2.$$
Experiments show that for small values of $k$
$$\exists x \in \mathbb{R}^{k+1} \,.\, \forall p \in [0,1] \,.\, S(p,x) = 0.$$
In other words, there are $x_i$'s such that $S(x,p)$ is identically zero as a polynomial in $p$.

For a given $k$ we can expand $S(x,p)$ as a polynomial in $p$ and equate the coefficients to $0$. For $k = 2$ we get \begin{align*} 0&=0 \\ -x_0^2-2 x_0+x_1^2&=0 \\ 2 x_0-2 x_1+1&=0 \\ \end{align*} and this has two solutions: $$x = (\frac{1}{2} (-1-\sqrt{2}),\frac{1}{2},\frac{1}{2} (3+\sqrt{2}))$$ and $$x = (\frac{1}{2} (-1+\sqrt{2}),\frac{1}{2},\frac{1}{2} (3-\sqrt{2})).$$ For $k = 1, 2, 3, 4, 5, 6, 7$ there are $1, 2, 4, 8, 14, 28, 48$ solutions respectively, according to Mathematica. According to OEIS this is A068912, "the number of $n$ step walks (each step $\pm 1$ starting from $0$) which are never more than $3$ or less than $-3$." This is kind of interesting because the problem arises in statistics, see John Mount's blog post for background.

**Question:** Is there a solution for every $k$?

**Addendum:** John says he wants soltions in $[0,1]^{k+1}$...

Here is the relevant Mathematica code:

```
s[k_, p_, x_] := Sum[Binomial[k, i] * p^i* (1 - p)^(k - i)* (Subscript[x, i] - p)^2, {i, 0, k}] Subscript[x, 0]^2
xs[k_] := Table[Subscript[x, i], {i, 0, k}]
system[k_, p_, x_] := Thread[CoefficientList[s[k, p, x], p] == 0]
solutions[k_] := Solve[system[k, p, x], xs[k], Reals]
```

To see the system of equations for $k = 4$, type

```
system[4, p, x] // ColumnForm
```

To see the solutions for $k = 4$, type

```
solutions[4]
```

To make a table of counts of solutions up to $k = 7$, type

```
Table[{k, Length@solutions[k]}, {k, 1, 7}] // ColumnForm
```