# Prove that $\gcd(f(x),g(x)) = 1$.

Let $F$ be a field, let $p , q$ be two coprime natural numbers and we consider the two polynomials $$f(X) = \sum_{i = 0}^{p - 1} X^{k i} \qquad \mbox{ and } \qquad g(X) = \sum_{j = 0}^{q - 1} X^{k j}$$ in $F[X]$, where $k \in \mathbb{N}$. If $x$ is an arbitrary element in $F$, how can I show that $\gcd(f(x) , g(x)) = 1$?

• If $F$ is a field, and $x\in F$, then aren't $f(x), g(x)$ just elements of $F$? Do you mean $\gcd(f, g)$ (alternatively $\gcd(f(X), g(X))$) instead? – Arthur Aug 20 '18 at 12:31
• Yes, if $x \in F$, then $f(x) , g(x)$ are in $F$. And I can alternatively show that $\gcd(f(X) , g(X)) = 1$ to end my proof. – joseabp91 Aug 20 '18 at 12:34

Hint (for the polynomial $\gcd$): $\;X^{kp} = \left(X^{k}-1\right)f(X)+1\,$ and $\,X^{kq} = \left(X^{k}-1\right)g(X)+1\,$.

Since $\,p,q\,$ are coprime, there exist integers $\,a,b\,$ such that $\,ap-bq=1\,$, and it can be assumed WLOG that they are positive i.e. $\,a,b \in \Bbb N\,$. Then $\,X^{kap}=X^{k(bq+1)} = X^k \cdot X^{kbq}\,$, and therefore:

$$\left(\left(X^{k}-1\right)f(X)+1\right)^{a} - X^k \cdot \left( \left(X^{k}-1\right)g(X)+1\right)^b = 0$$

Expanding the binomials, $\,h(X) = \gcd\left(f(X),g(X)\right)\,$ would have to divide $\,1-X^k\,$, but both $\,f(X)\,$ and $\,g(X)\,$ are coprime with $\,1-X^k\,$.

• Expanding $$\left(\left(X^{k}-1\right)f(X)+1\right)^{a} - X^k \cdot \left( \left(X^{k}-1\right)g(X)+1\right)^b$$ I got that there exists $h(X) \in F[X]$ such that $$(X^k - 1) h(X) = \left(\left(X^{k}-1\right)f(X)+1\right)^{a} - X^k \cdot \left( \left(X^{k}-1\right)g(X)+1\right)^b = 0\mbox{.}$$ Why do you say that $\gcd(f(X) , g(X))$ would have to divide $1 - X^k$? And why are both $f(X)$ and $g(X)$ coprime with $1 - X^k$? – joseabp91 Aug 21 '18 at 21:08
• @joseabp91 See the detailed proof for $k=1$ here. The case $k \gt 1$ works out in an entirely similar way. – dxiv Aug 21 '18 at 21:58
• Thank you very much by your anwers. – joseabp91 Aug 21 '18 at 22:25

Perhaps this will help:

$$1+x+x^2+...+x^{n-1} = {x^n-1\over x-1}$$

and a fact that $$\gcd(x^n-1,x^m-1) = x^{\gcd(m,n)}-1$$

• It is not valid for me, as my statement and the fact $\gcd(x^n - 1 , x^m - 1) = x^{\gcd(n , m)} - 1$ are clearly equivalent. Unless you help me to prove your statement. – joseabp91 Aug 20 '18 at 12:36
• I'm almost sure you can find a proove on this site. Perhaps even in my posts you can find it. – Aqua Aug 20 '18 at 12:37