# Reason for substitution : $a=z + \frac{1}{z}$.

Have read regarding the substitution $a = z + \frac{1}{z}$ to get the factorization of $z^6+z^5+z^4+z^3 +z^2+z+1$ to get the form $a^3+a^2-2a -1=0$ in the book by Erickson martin, titled: Beautiful mathematics, on page #$58$ as shown below. I am unable to get the process for the division, i.e. how to divide by $a$ the given polynomial. I mean that $a = \frac{z^2+1}{z}$ cannot divide $z^6+z^5+z^4+z^3 +z^2+z+1$.

For $z^5-1=0\implies (z-1)(z^4+z^3+z^2+z+1)=0$, want to use the same logic of symmetry for $z+z^4= z+\frac1z=a'$, but am hindered by the inability to divide $(z^4+z^3+z^2+z+1)$ by $a'$.

• Commented May 16, 2018 at 6:12
• @labbhattacharjee Thanks a ton. Commented May 16, 2018 at 6:20
• See this old thread. Commented Jul 22, 2023 at 11:21

We say that a polynomial is palindromic, if its sequence of coefficients can equally well be read backwards. So a degree $$n$$ polynomial $$p(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n,\ a_n\neq0,$$ is palindromic if $$a_i=a_{n-i}$$ for all indices $$i, 0\le i\le n$$. This system of equations can compactly be restated in the form $$p(x)=x^np(\frac1x).$$ This way of doing it shows that $$p(\alpha)=0$$ if and only if $$p(1/\alpha)=0$$.

If we further assume that $$n$$ is even, say $$n=2k$$, then we get to the business part. In that case we can write $$\frac1{x^k}p(x)=a_0x^{-k}+a_1x^{-k+1}+\cdots+a_{k-1}x^{-1}+a_k+a_{k+1}x+\cdots a_{2k}x^k.\qquad(*)$$ Observe that here the coefficients of $$x^i$$ and $$x^{-i}$$ are equal as a consequence of the palindromic property. This means that $$(*)$$ can be written as a polynomial in the new variable $$z=x+\dfrac1x$$. Behold \begin{aligned} x+\frac1x&=z,\\ x^2+\frac1{x^2}&=(x+\frac1x)^2-2=z^2-2,\\ x^3+\frac1{x^3}&=(x^2+\frac1{x^2})(x+\frac1x)-(x+\frac1x)=z^3-3z,\\ \vdots\\ x^{\ell+1}+\frac1{x^{\ell+1}}&=(x^\ell+\frac1{x^\ell})(x+\frac1x)-(x^{\ell-1}+\frac1{x^{\ell-1}})=\cdots.\\ \end{aligned} Here the last line contains a general recurrence relation defining a sequence of polynomials $$q_\ell, \ell=1,2,\ldots$$, such that $$x^\ell+x^{-\ell}=q_\ell(z)$$. We simply declare $$q_0(z)=2, q_1(z)=z$$, and apply the recurrence $$q_{\ell+1}(z)=z q_\ell(z)-q_{\ell-1}(z)$$ for higher values of $$\ell$$.

Therefore the right hand side of $$(*)$$ is equal to $$a_k+a_{k-1}q_1(z)+a_{k-2}q_2(z)+\cdots+a_0q_k(z)=\sum_{i=0}^ka_{k-i}q_i(z).\qquad(**)$$

Observe that $$\deg q_i=i$$, so on the right hand side of $$(**)$$ we have a degree $$k$$ polynomial.

What all this implies is that we can find the zeros of a palindromic polynomial $$p(x)$$ of even degree $$n=2k$$ by the process of:

1. Write $$x^k(p(1/x)$$ in the form $$f(z)$$ with a polynomial $$f$$ of degree $$k$$.
2. Find the zeros of $$f(z)$$ (this may still be taxing if $$k$$ is large).
3. For each zero $$z_j, j=1,2,\ldots,k,$$ solve the quadratic equation $$x+\frac1x=z_j$$ to find two of the zeros of $$p(x)$$.

Example. When $$p(x)=x^4+x^3+x^2+x+1$$, a palindromic of degree $$4=2\cdot2$$, we see that $$x^2p(\frac1x)=x^2+x+1+\frac1x+\frac1{x^2}=1+q_1(z)+q_2(z)=z^2+z-1.$$ The zeros of $$z^2+z-1$$ are $$z_{1,2}=(-1\pm\sqrt5)/2$$. The rest is easy.

• For a consequence in Galois theory see this thread. Commented May 16, 2018 at 6:33
• For a modern, understandable description of the way that Gauss did this, zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf with a ton of examples at books.google.com/… Commented May 16, 2018 at 14:41
• +1 It is amazing how you explain things. You should write a book, I will be the first one to buy it. Commented Feb 7, 2019 at 7:52

I don't understand why you are referring to division. All you have to do is substitute $$a=z+\frac1z$$ into $$a^3+a^2-2a-1$$ and simplify. You should get $$\frac{z^6+z^5+z^4+z^3+z^2+z+1}{z^3}\ .$$ (I believe the $z^6+z^5+z^4+z^3+z^2+1$ in your question is wrong.) Then you have $$z^6+z^5+z^4+z^3+z^2+z+1=0\quad\hbox{if and only if}\quad a^3+a^2-2a-1=0\ .$$

• But, how to get the polynomial $a^3 +a^2 -2a -1$ is the key. I am unable to derive such polynomial for my case of $z^4 +z^3+z^2 +z +1$. Commented May 16, 2018 at 6:16
• See Jyrki Lahtonen's answer. Commented May 16, 2018 at 6:49

$$a=z+\dfrac{1}{z}\Rightarrow a^2=\left(z+\dfrac{1}{z}\right)^2=z^2+\dfrac{1}{z^2}+2$$

$$a=z+\dfrac{1}{z}\Rightarrow a^3=\left(z+\dfrac{1}{z}\right)^3=z^3+3\dfrac{z^2}{z}+3\dfrac{z}{z^2}+\dfrac{1}{z^3}=z^3+\dfrac{1}{z^3}+3z+\dfrac{3}{z}$$

$$a=z+\dfrac{1}{z}\Rightarrow -2a=-2z-\dfrac{2}{z}$$

Then we will have

$$a^3+a^2-2a-1=0$$

$$\Leftrightarrow z^3+\dfrac{1}{z^3}+3z+\dfrac{3}{z}+z^2+\dfrac{1}{z^2}+2-2z-\dfrac{2}{z}-1=0$$

$$\Leftrightarrow z^3+x^2+z+1+\dfrac{1}{z}+\dfrac{1}{z^2}+\dfrac{1}{z^3}=0$$

$$\Leftrightarrow z^3\left(z^3+x^2+z+1+\dfrac{1}{z}+\dfrac{1}{z^2}+\dfrac{1}{z^3}\right)=0$$

$$\Leftrightarrow z^6+z^5+z^4+z^3+z^2+z+1=0$$

If this problem appears as an exercise, it should be done like this (reverse the steps above):

$$z^6+z^5+z^4+z^3+z^2+z+1=0$$, we know that $$z=0$$ is not a root of this equation

$$\Leftrightarrow \dfrac{z^6+z^5+z^4+z^3+z^2+z+1}{z^3}=0$$

$$\Leftrightarrow z^3+z^2+z+1+\dfrac{1}{z}+\dfrac{1}{z^2}+\dfrac{1}{z^3}=0$$

$$\Leftrightarrow z^3+\dfrac{1}{z^3}+3z+\dfrac{3}{z}+z^2+\dfrac{1}{z^2}+2-2z-\dfrac{2}{z}-1=0$$

Let $$a=z+\dfrac{1}{z}$$, we have:

$$a=z+\dfrac{1}{z}\Rightarrow a^2=\left(z+\dfrac{1}{z}\right)^2=z^2+\dfrac{1}{z^2}+2$$

$$a=z+\dfrac{1}{z}\Rightarrow a^3=\left(z+\dfrac{1}{z}\right)^3=z^3+3\dfrac{z^2}{z}+3\dfrac{z}{z^2}+\dfrac{1}{z^3}=z^3+\dfrac{1}{z^3}+3z+\dfrac{3}{z}$$

$$a=z+\dfrac{1}{z}\Rightarrow -2a=-2z-\dfrac{2}{z}$$

then $$a^3+a^2-2a-1=0$$.