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let $f(x)=x^3-1$ and $g(x)=a_0+a_1x+a_2x^2$, where $f(x),g(x) \in \mathbb{Z}[x]$, and $\textrm{GCD}(f(x),g(x))=1$. From all this, is it possible to infer anything about the resultant $\textrm{Res}(f(x),g(x))$? I know it has to be non-zero and also $a_0+a_1+a_2 =1$ (i proved this bit using other properties).but I was wondering if there was any other property the resultant had. I ask because i eventually want to solve the diophantine equation $\textrm{Res}(f(x),g(x))= 1$, this equation is not easy to solve as n increases so i was hopping there were other properties of the resultant i could exploit to make the equation simpler

Thanks in advance

eagle I
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2 Answers2

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let $f(x)=x^3-1, g(x)=a_0+a_1x+a_2x^2$, $f(x),g(x) \in \mathbb{Z}[x]$, and $GCD(f(x),g(x))=1$. From all this, is it possible to infer anything about the resultant $Res(f(x),g(x))$? I know it has to be non-zero

Indeed, $\,\operatorname{Res}(f(x),g(x)) \ne 0\,$ follows from $\,\gcd(f(x),g(x))=1\,$, because the resultant is $\,0\,$ iff the two polynomials have at least one common root.

and also $\,a_0+a_1+a_2 \neq 0\,$ (otherwise $\,\operatorname{Res}(f(x),g(x))=0\,$)

The condition also follows from $\,\gcd(f(x),g(x))=1\,$, because $\,a_0+a_1+a_2 = 0$ $\iff \,g(1) = 0\,$, and in that case $\,(x-1) \mid \gcd(f(x),g(x))\,$ since $\,1\,$ is also a root of $\,f(x) = x^3 -1\,$.

By the same $\,\gcd\,$ condition, the complex cube roots of unity, which are the other two roots of $\,f(x)\,$, must not also be roots of $\,g(x)\,$, which is equivalent to $\,a_0, a_1, a_2\,$ not being all three equal.

but i was wondering if there was any other property the resultant had. I ask because i eventually want to solve the equation $Res(f(x),g(x))= 1$

It is not clear what other properties you are looking for, or what unknown(s) the equation you mention at the end would be solved for. But the resultant can be calculated explicitly, so you can look at it and assess yourself whether it has the desired properties. This can be done either with the help of a CAS (for example, WA), or directly by hand (see e.g. 1, 2), and the result is:

$$ \begin{align} \operatorname{Res}(f(x),g(x)) &= a_0^3 + a_1^3 + a_2^3 - 3 a_0 a_1 a_2 \\ &= (a_0 + a_1 + a_2) (a_0^2 + a_1^2 + a_2^2 - a_0 a_1 - a_0 a_2 - a_1 a_2) \\ &= \frac{1}{2}(a_0 + a_1 + a_2) \big((a_0-a_1)^2 + (a_0-a_2)^2 + (a_1-a_2)^2\big) \end{align} $$

dxiv
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  • Thanks for the detailed response. sorry for not making it clear, by other property i meant a property that would make solving the diophantine equation $Res(f(x),g(x)) = a_0^3+a_1^3+a_2^3-3a_0a_1a_2 = 1$ for the $a_i$'s easier. I somehow feel like there is something special about the resultant being 1 and the polynomials being co-prime and was hopping that could be exploited somehow. – eagle I Jun 10 '23 at 05:33
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    @eagleI From the above you get $a_0 + a_1 + a_2$ $= a_0^2 + a_1^2 + a_2^2 - a_0 a_1 - a_0 a_2 - a_1 a_2 = 1$ for the diophantine equation, then the solution is immediate if you are looking for non-negative integers, otherwise see Help with a specific Diophantine Equation. – dxiv Jun 10 '23 at 06:03
  • thanks for the link. But do you think there is anything special about the polynomials being co-prime and the resultant being 1 that would make the diophantine equation simpler? i ask because i want to eventually generalize this and solve the equation $Res(x^n-1,a_0+a_1x+a_2x....a_{n-1}x^{n-1}) = 1$. where $n \in \mathbb{N}$ So i was hopping the properties i stated would make solving the general diophantine equation simpler. If not, do you know of any way to solve the general diophantine equation (can be papers, topics/areas of math i could search up, anything really)? – eagle I Jun 10 '23 at 06:24
  • @eagleI You could still deduce that $\sum a_i \ne 0$ and, at least in the case of a prime $n$, that not all $a_i$ can be equal. Don't know that much else carries over from the $n=3$ case, though I haven't given it a lot of thought, honestly. It might help if you elaborated some more on the context. I don't suppose you just take resultants of arbitrary polynomials on a whim and study the associated diophantine equations. – dxiv Jun 10 '23 at 07:03
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    actually ∑=1 (proved this using another method). As for the context, i was studying the units of the ring $\mathbb{Z}[x]/<x^n-1>$ (note, i am aware that this is isomorphic to the group ring $\mathbb{Z}[C_n]$) and i was able to prove that finding units of that ring corresponds to solving the diophantine equation involving the resultant of $f(x)=x^n-1$ and $g(x)=a_0+a_1x+a_2x^2....+a_{n-1}x^{n-1}$. I was able to put more restrictions on this and was able to prove $GCD(f(x),g(x))=1$. i was hopping studying this diophantine equation would be simpler but apparently not :( – eagle I Jun 10 '23 at 07:32
  • @eagleI Thanks for the background. This sounds related to, though not the same as this. You could post it as a separate question with all the necessary context, and maybe add a number-theory tag, too. – dxiv Jun 10 '23 at 18:11
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The resultant is get using the old Sylvester's method. We have $$x^3+0x^2+0x-1=0\\a_2x^2+a_1x+a_0=0$$Then the resultant is an homogenous function, rational, integer, of $3$ degree respect to the coefficients of second polynomial and of $2$ degree respect to those of the first one.

$$\begin{vmatrix}1&0&0&-1&0\\0&1&0&0&-1\\a_2&a_1&a_0&0&0\\0&a_2&a_1&a_0&0\\0&0&a_2&a_1&a_0\end{vmatrix}=0$$ So you have $$a_2^3+a_1^3+a_0^3-3a_2a_1a_0=0$$

I think the other properties you want should be deduced, (if any interesting and useful) from this resultant itself.

Piquito
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