# 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;
$n > 0$.

• Might you have meant $a\in[-1,1]$ or "$a$ is in $[-1,1]$" where you wrote "$a$ is $[-1,1]$"? ${}\qquad{}$ – Michael Hardy May 7 '15 at 15:06
• The first option – Sofia May 7 '15 at 15:08
• @Sofia You didn't have to choose: $a\in[-1,1]$ and $a$ is in $[-1,1]$ means the same. – ajotatxe May 7 '15 at 15:10
• @MichaelHardy: $a\in[-1,1]$ means the same as "$a$ is in $[0,1]$". I suspect the OP meant to say "$a_i \in \{-1,1\}$". At least that makes the statement true! – TonyK May 7 '15 at 15:10
• @TonyK : I now suspect you're right. Six answers have appeared so far, and not all of them construe the question that way. ${}\qquad{}$ – Michael Hardy May 7 '15 at 15:16

What is true is that if $a_n = 1$, $p \ge 2$ and all $a_i \in [-1,1]$, then $\sum_{i=0}^n a_i p^i > 0$.

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.

$$0.01p-0.03=0$$
No. $p=2$, $n = 1$, $a_0 = 1$ and $a_1 = -\frac 12$ gives $$a_0 + a_1p^1 = 1 -\frac 12 \cdot 2 = 0.$$
Let $n=1$, $a_n=-\frac1p$, $a_0=1$.
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 $\pm 1$.