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Let $P (x)$ be a polynomial with integer coefficients. It is known that the numbers $1$ and $2$ are its roots. Prove that there exists a coefficient that is less than $-1$.

My work so far:

Let $P(x)=a_nx^n+...+ax+a_0$. $P(1)=P(2)=0$. Prove that exists $a_i<-1$.

Suppose $\forall a_i\ge-1$. Let $S$ be the sum of positive factors. If $P(1)=0$ then $a_n+..+a_1+a_0=0$, then number of negative factors equally $S$.

I need help here.

Najib Idrissi
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Roman83
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1 Answers1

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We assume that $deg P(x) \geq 2$ and that $a_n \neq 0$. Since $P(1) = P(2) = 0$, we have $a_n2^n + a_{n - 1}2^{n-1} + .. + a_12 + a_0 = 0$ and $a_n + a_{n - 1} + .. + a_1 + a_0 = 0$. Together, we have $(a_{n - 1} + .. + a_1 + a_0)2^n = a_{n - 1}2^{n - 1} + .. + a_12 + a_0$. If we assume that $a_k \geq -1$ for $k \leq n - 1$, then $a_{n - 1}2^{n - 1} + .. + 2a_1 + a_0 > -2^n$. Therefore $(a_{n - 1} + .. + a_1 + a_0)2^n > -2^n$; moreover, $a_{n-1} + .. + a_1 + a_0 > -1$ which implies that $a_n < 1$. Since $a_n \neq 0$, we have $a_n \leq -1$. However, if $a_n = -1$, then $a_{n-1} + .. + a_1 + a_0 = 1$ with $a_k \geq -1$ for $k \leq n - 1$ (as well as $n - 1 \geq 1$ because of degree of $P(x)$) and $a_{n-1}2^{n-1} + .. + a_12 + a_0 = 2^n$ which leads to a contradiction, just observe the two cases for $n \geq 3$: (1) $a_{n-1} = 2$ and $a_0 = -1$; (2) $a_{n - 1} = 3$ and $a_{n-2} = a_{n-3} = -1$.

Steven Harding
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