a) If $\alpha$ is root of the polynomial $p(x)=a_0+a_1x+\dots+a_nx^n$ with real coefficients, $a_n\neq 0$, then prove that $|\alpha|\le 1+\max_{ 0\le k\le n-1}\left|\frac{a_k}{a_n}\right|$.

b) Let $a_0+10a_1+\dots+10^na_n$ be the decimal representation of a prime number such that $a_n\ge 2,n>1$. Prove that the polynomial $p(x)=a_0+a_1x+\dots+a_nx^n$ cannot be written as a product of two non-constant polynomials with integer coefficients.

I don't know how i prove the first part. I use Vieta's formula but cannot prove it and the second one i think i have to prove by contradiction method.I assume that $f(x)$ can be written as a product of two polynomial and after that i equate the coefficients but then i unable to prove the contradiction.

  • $\begingroup$ You should split that into two questions. $\endgroup$ – Macavity Jan 7 '18 at 16:41
  • $\begingroup$ @RAM_3R Is $a_n\leq 2$ or $a_n\geq 2$? $\endgroup$ – Robert Z Jan 7 '18 at 17:59
  • $\begingroup$ sry typo , you're right.. $\endgroup$ – RAM_3R Jan 7 '18 at 18:01

As regards a) the stronger inequality $|\alpha|<1+R$ holds where $\alpha$ is a root of $p$ and $R:=\max_{ 0\le k\le n-1}\left|\frac{a_k}{a_n}\right|$.

If $R=0$ then $\alpha=0$ and, trivially, $|\alpha|<1+R$. Assume now that $R>0$. If $|\alpha|\leq 1$ then $|\alpha|<1+R$. If $|\alpha|>1$ then $$\begin{align} |\alpha|^n&\leq \frac{|a_0|}{|a_n|}+\frac{|a_1||\alpha|}{|a_n|}+\dots +\frac{|a_{n-1}||\alpha|^{n-1}}{|a_n|}\\ &\leq R(1+|\alpha|+\dots +|\alpha|^{n-1})=R\frac{|\alpha|^n-1}{|\alpha|-1}<R\frac{|\alpha|^n}{|\alpha|-1}\end{align}$$ which implies that $|\alpha|^{n+1}-|\alpha|^{n}< R|\alpha|^n$, that is $|\alpha|< 1+R$.

For b), note that $1+R:=1+\max_{ 0\le k\le n-1}\left|\frac{a_k}{a_n}\right|\leq 1+9/2=11/2$. Hence $|\alpha|<11/2$ for any root $\alpha$ of $p$. Now assume that $p=AB$ with $A$ and $B$ non-constant polynomials in $\mathbb{Z}[x]$. If $p(10)$ is a prime then $|A(10)|=1$ or $|B(10)|=1$, but $$1=\left|c\prod_{k=1}^m (10-\alpha_k)\right|\geq \prod_{k=1}^m (10-|\alpha_k|)>\prod_{k=1}^m(10-11/2)>1.$$ which is a contradiction.


For part b, a hint. If $p$ can be written as a product of two polynomials, what does that say about $p(10)$, your prime?

  • 1
    $\begingroup$ ohhh i got it .$p(x)=g(x)h(x)$Now $p(10)=g(10)h(10)$but left side a prime no ....is it right?? $\endgroup$ – RAM_3R Jan 7 '18 at 16:47

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