I have to prove that:
If $a$ and $b$ are two roots of $x^4+x^3-1=0$, then $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$.
I tried this :
$a$ and $b$ are root of $x^4+x^3-1=0$ means :
$\begin{cases} a^4+a^3-1=0\\ b^4+b^3-1=0 \end{cases}$
which gives us :
$(ab)^4+(ab)^3=a^3+b^3+a^4+b^4+a^4b^3-a^3b^4-1$
can you help me carry on ? or propose another solution ? thanks in advance
Let $a,b,c,d$ be the roots of $x^4+x^3-1=0$.
By Vieta's formula, $$a+b+c+d=-1\quad\Rightarrow\quad c+d=-1-(a+b)\tag1$$ $$abcd=-1\quad\Rightarrow \quad cd=-\frac{1}{ab}\tag2$$
Since we have $$a^4+a^3=1\quad\text{and}\quad b^4+b^3=1$$ we can have $$1=(a^4+a^3)(b^4+b^3)$$ $$(ab)^4+(ab)^3(a+b+1)=1,$$ i.e. $$a+b=\frac{1-(ab)^4}{(ab)^3}-1\tag3$$ Similarly, $$(cd)^4+(cd)^3(c+d+1)=1$$ Using $(1)(2)$, this can be written as $$\left(-\frac{1}{ab}\right)^4+\left(-\frac{1}{ab}\right)^3(-1-(a+b)+1)=1,$$ i.e. $$a+b=\left(1-\frac{1}{(ab)^4}\right)(ab)^3\tag4$$ From $(3)(4)$, letting $ab=x$, we have $$\frac{1-x^4}{x^3}-1=\left(1-\frac{1}{x^4}\right)x^3$$ to get $$x^6+x^4+x^3-x^2-1=0.$$
Lemma: If $\alpha,\beta,\gamma$ are the roots of an irreducible polynomial over $\mathbb{Q}$ with degree $3$, $\alpha\beta$ is a root of: $$ q(x)=(x-\alpha\beta)(x-\alpha\gamma)(x-\beta\gamma)\in\mathbb{Q}[x]. $$
Proof: The coefficients of $q(x)$ are symmetric polynomials in $\alpha,\beta,\gamma$, hence $q(x)\in\mathbb{Q}[x]$.
The same argument also shows that:
Lemma: If $\alpha,\beta,\gamma,\delta$ are the roots of an irreducible polynomial over $\mathbb{Q}$ with degree $4$, $\alpha\beta$ is a root of: $$ q(x)=(x-\alpha\beta)(x-\alpha\gamma)(x-\beta\gamma)(x-\alpha\delta)(x-\beta\delta)(x-\gamma\delta)\in\mathbb{Q}[x]. $$
By applying Viète's formulas to the polynomial $x^4+x^3-1$, we have: $$\alpha+\beta+\gamma+\delta = -1, $$ $$\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=0, $$ $$\alpha\beta\gamma\delta = -1,$$ $$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}=\frac{1}{\alpha\beta}+\frac{1}{\alpha\gamma}+\frac{1}{\alpha\delta}+\frac{1}{\beta\gamma}+\frac{1}{\beta\delta}+\frac{1}{\gamma\delta}=0,$$ so the elementary symmetric polynomials of $\alpha\beta,\alpha\gamma,\alpha\delta,\beta\gamma,\beta\delta,\gamma\delta$ are straightforward to compute, giving $q(x)=x^6+x^4+x^3-x^2-1$ as wanted.
The companion matrix of $p(z) = z^4 + z^3 - 1$ is $$ A = \pmatrix{0 & 0 & 0 & 1\cr 1 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & -1\cr}$$ This is a matrix whose eigenvalues are the roots of $p(z)$. The Kronecker product $A \otimes A$ is a $16 \times 16$ matrix; all products $ab$ where $a$ and $b$ are roots of $p(z)$ are eigenvalues of $A \otimes A$. Moreover, if $a \ne b$, $ab$ will be an eigenvalue of multiplicity $\ge 2$, because if $u$ and $v$ are eigenvectors of $A$ for eigenvalues $a$ and $b$, both $u \otimes v$ and $v \otimes u$ are eigenvectors of $A \otimes A$ for eigenvalue $ab$.
Now the characteristic polynomial of $A \otimes A$ turns out to be $$\left( {z}^{4}-{z}^{3}-2\,{z}^{2}+1 \right) \left( {z}^{6}+{z}^{4}+{ z}^{3}-{z}^{2}-1 \right) ^{2} $$ The products $ab$ with $a \ne b$ are roots of this characteristic polynomial with multiplicity $\ge 2$, and therefore are roots of $ {z}^{6}+{z}^{4}+{ z}^{3}-{z}^{2}-1 $.
In the case $a=b$, $ab = a^2$ turns out to be a root of the other factor $ {z}^{4}-{z}^{3}-2\,{z}^{2}+1 $.
