My problem is the following:
1) gcd(a,b)=1. I need to prove that $$\gcd(a-b, c_0a^n+c_1a^{n-1}b+...+c_{n-1}ab^{n-1}+c_nb^n)=\gcd(a-b, c_0+c_1+...c_n).$$ I know that $x \equiv y$ (mod m) implies $\gcd(m, x)=\gcd(m, y)$. So I am looking to rewrite $c_0a^n+c_1a^{n-1}b+...+c_{n-1}ab^{n-1}+c_nb^n$ and to put it in the following form $(a-b)*(...)+(c_0+c_1+...c_n)$. Can you help me?
2) Also, using $a=10$ and $b=1$, I need to "derive the test for divisibility by 9" from the previous formula.
Thank you.
Hint $ $ By Euclid $\ \gcd(a\!-\!b,f(a,b)) = \gcd(a\!-\!b, f(a,\color{#c00}b)\bmod a\!-\!b) = \gcd(a\!-\!b,f(a,\color{#c00}a))\ $ for any polynomial $\,f\,$ with integer coefficients, by applying $\,\color{#c00}{b\equiv a}\pmod{\!a\!-\!b}$