Denominator in rational gcd of integer polynomials

Solution 1:

• Regarding Question $1$: I think that $d$ in your post is -in the general case- some divisor of the resultant. However, any divisor $r>1$ of the resultant would do, in the sense that it may be the $\gcd\big(f(n), g(n)\big)$ for some $n$:

for any divisor $r$ of the resultant $Res(f,g)=d$, we can always find some integer $n$ such that $\gcd\big(f(n),g(n)\big)\geq r>1$, given that the following system of simultaneous equations $$ \begin{array}{c} f(x)\equiv 0\mod r \\ % \\ g(x)\equiv 0\mod r \end{array} $$ has a solution $x\equiv n\mod r$.

Then, the secuence $$ m_{n,r}(i)=n+r*i $$ for all $i\geq0$, has the property that: $$ \gcd\Big(f\big(m_{n,r}(i)\big),g\big(m_{n,r}(i)\big)\Big)\geq r>1 $$

• Regarding Question $2$: The resultant $Res(f,g)$ of two polynomials over a field, is by definition the determinant of the Sylvester matrix of the polynomials, and it can be shown to be equal to the product between the differences of their roots (some of them may lie in some field extension) times suitable powers of the leading coefficients of $f,g$.

• Regarding Question $3$: The answer is negative in general:

For example consider the polynomials $f(x)=5x+1$, $g(x)=5x+6$. Then $Res(f,g)=25$ but the simultaneous equations $$ \begin{array}{c} 5x+1\equiv 0\mod 5 \\ % \\ 5x+6\equiv 0\mod 5 \end{array} $$ have no common solutions: $f-g=5$ thus any common divisor of $f(n), g(n)$ must divide $5$. However, none of $f(n), g(n)$ is divisible by $5$. Thus, $\gcd(5n+1,5n+6)=1$ for any positive integer $n$. Notice also that $$ u(x)f(x)+w(x)g(x)=5 $$ for $u(x)=-1$ and $w(x)=1$.