Polynomial with a prime number as a root Is it possible to prove that this equation is false:
$$
\sum_{i=0}^n a_i p^i = 0
$$
with following conditions:
$a_i \in [-1;1]$; [Might $a\in\{-1,1\}$ have been intended here?] 
$p$ is a prime number; 
$n > 0$.
 A: I assume you meant $a_i \in \{-1,1\}$. The same argument works if $ a_n = \pm 1$ and the rest of the $a_i$'s are in the interval $[-1,1]$ or even more generally if $\vert a_ n\vert \geq \vert a_i \vert$ for all $i \in \{0,1,2\ldots,n-1\}$.
If so, assume what you have is true and we will obtain a contradiction.
We have
$$\sum_{i=0}^{n-1} a_i p^i = - a_n p^n$$
This gives us
$$\left \vert \sum_{i=0}^{n-1} a_i p^i \right \vert = \left \vert - a_n p^n \right \vert = p^n$$
since $\vert -a_n \vert = 1$. We have
$$p^n = \left \vert \sum_{i=0}^{n-1} a_i p^i \right \vert \leq \sum_{i=0}^{n-1} \left \vert a_i p^i\right \vert = \sum_{i=0}^{n-1} p^i = \dfrac{p^n-1}{p-1} \leq p^n-1$$
which gives us a contradiction.
A: How about this?
$$0.01p-0.03=0$$
A: What is true is that if $a_n = 1$, $p \ge 2$ and all $a_i \in [-1,1]$, then
$\sum_{i=0}^n a_i p^i > 0$.
A: No. $p=2$, $n = 1$, $a_0 = 1$ and $a_1 = -\frac 12$ gives 
$$ a_0 + a_1p^1 = 1 -\frac 12 \cdot 2 = 0. $$
A: Let $n=1$, $a_n=-\frac1p$, $a_0=1$.
A: If you mean $a_i \in \{-1, 1\}$ then you can just use the rational root's test to show that the only possible rational roots are $\pm 1$.
