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

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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Yes, it is true, but you should still clarify if you regard the gcd of polynomials or for a particular value of $x$. – Phira Oct 31 '11 at 11:04
@Phira,gcd of polynomials...by the way: en.wikipedia.org/wiki/… – pedja Oct 31 '11 at 11:09

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

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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$a+1$ and $b+1$ are coprime...not necessarily $a$ and $b$ – pedja Oct 31 '11 at 11:35
What do you think "note the shift!" means? :-) – Phira Oct 31 '11 at 11:36
I am not sure what you mean with divisiblity theorem. Your particular polynomials are products of cyclotomic polynomials, and there are many ways to see your result or my generalization by looking at the roots of unity or citing results on cyclotomic polynomials. – Phira Oct 31 '11 at 11:54
@pedja: Yes, of course, since $f_n$ is not 1 for $n$ greater than 1. Actually, the reverse direction is simpler by factorizing $x^k-1$. – Phira Oct 31 '11 at 13:44
@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. – Phira Oct 31 '11 at 14:53