Construct a polynomial with $k$ values of $1$ and $-1$

Question

How can I construct a polynomial with $k$ different integers $\alpha _k$, such that $f(\alpha )=1$ or $f(\alpha )=-1$

$$\begin{align*}f(x)=\left(x-\alpha _1\right)\left(x-\alpha _1\right)\text{...}\left(x-\alpha _n\right)+1\tag{1}\end{align*}$$

$$\begin{align*}f(x)=\left(x-\alpha _1\right)\left(x-\alpha _1\right)\text{...}\left(x-\alpha _k\right)+(-1)^{\text{function}}\tag{2}\end{align*}$$

Answer 1

In general, given $\alpha_1, \dotsc, \alpha_n$, let $$p_k (x): = \frac{\prod_{j \neq k }(x-\alpha_j)} {\prod_{j\neq k}(\alpha_k - \alpha_j)}.$$

Then $\sum_{k=1}^na_kp_k(x)$ will have values $a_k$ at $\alpha_k$.

Answer 2

What about

$$f(x):=(x-\alpha)(x-\alpha_1)\cdot\ldots\cdot(x-\alpha_k)\pm1\;\ldots$$

Now just choose the sign.