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Let us define :

$P(x)=x^{a}+x^{a-1}+\cdots+1$

$Q(x)=x^{b}+x^{b-1}+\cdots+1$ , where $a>b$ and $a , b \in \mathbb{N}$

Is it true that:

If $\gcd(a+1,b+1)=1 \Rightarrow \gcd(P(x),Q(x))=1$ ?

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    @Phira,gcd of polynomials...by the way: http://en.wikipedia.org/wiki/Greatest_common_divisor_of_two_polynomials#Formal_definition2011-10-31

1 Answers 1

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$P(x)-x^{a-b}Q(x)=x^{a-b-1}+\dots+1$.

Set $f_n=x^{n-1}+\dots+1$ (note the shift!).

So you know that $\gcd(f_a,f_b)=\gcd(f_b,f_{a-b})$.

But this is exactly the key relation for calculating the gcd of two numbers.

So $\gcd(f_a,f_b)=f_{\gcd(a,b)}.$

In particular, if $a$ and $b$ are coprime, we get $f_1$ which is 1. (Since we only multiplied by powers of $x$ this is true on the level of polynomials as well as on the level of particular values of $x$.)

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    @pedja Why is it better if Yimin wrote it as$a$pdf than if I wrote it here in mathjax? This fact is typically an exercice in standard number theory or algebra textbooks, but of course, I do not remember for sure in which book or which exercice number you can find it.2011-10-31