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