Can a polynomial of the form $\sum\limits_{k=0}^n a_k x^k$, where $a_k=\pm1$, have arbitrarily many distinct real roots?
I suspect the answer is yes. So far, the most roots of such a polynomial that I have found, is five: $x^{13}+x^{12}-x^{11}-x^{10}-x^9-x^8-x^7+x^6+x^5-x^4+x^3-x^2+x+1$.
(This question was inspired by a question, "What is the maximum number of distinct real roots of a polynomial whose coefficients are all $1$, $-1$ or $0$?")
Best Answer
From Sil's answer to another question, and Calvin Lin's comment:
Let $f(x)=x^2-x-1$.
The polynomial $g(x)=f(x)f(x^3)f(x^9)...f(x^{(3^m)})$ is of the form $\sum\limits_{k=0}^n a_k x^k$ where $a_k=\pm1$.
By increasing $m$, $g(x)$ can have arbitrarily many distinct real roots (as proved by Sil's answer).