If $f$ is an integer coefficient polynomial and $n$ is square free, does $f(\frac{1+\sqrt n}{2})=f(\frac{1-\sqrt n}{2})$ imply that $f$ is an integer?

Question

My question:

If $f$ is an integer coefficient polynomial and $n$ is a square free integer, does $f(\frac{1+\sqrt n}{2})=f(\frac{1-\sqrt n}{2})$ imply that $f$ is an integer?

Suppose $\sum_ia_i(\frac{1+\sqrt n}{2})^i=\sum_ia_i(\frac{1-\sqrt n}{2})^i$, where $a_i$ are integers. I wonder if we can conclude that $a_i=0$ except for $a_0$ from here.

Answer 1

If $f(x) =(x-\frac{1+\sqrt n}{2})(x-\frac{1-\sqrt n}{2}) =x^2-x+\frac{1-n}{4} $ then $f(\frac{1+\sqrt n}{2})=f(\frac{1-\sqrt n}{2}) =0 $ and if $n=4m+1$ then $f(x)$ has integer coefficients.