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This is Exercise 14 on page 110 of Analysis I by Amann and Escher.

The hint given is as follows: multiply the polynomial by $1/X^2$ and substitute $Y = X - 1/X$.

If I attempt this, I get the following:

\begin{align*} X^4 - 2X^3 - X^2 + 2X + 1 &= 0\\ \Rightarrow X^2 - 2X - 1 + \frac{2}{X} + \frac{1}{X^2} &= 0. \end{align*}

My problem is that I don't understand how to make the suggested substitution. I'm wondering if there is something obvious I'm missing.

I appreciate any help.

Michael Rozenberg
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Novice
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4 Answers4

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$\left(x-1/x\right)^2=x^2+1/x^2-2$. If you substitute $y=x-1/x$, your equation becomes $y^2+2-2y-1=0$, so $y^2-2y+1=0$, so $y=1$. Then you find $x$ from $x-1/x=1$: $x^2-x-1=0$, $x=(1\pm \sqrt{5})/2$. Each of these is a double root of the original equation.

markvs
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$y^2=x^2-2+1/x^2$, so the expression after multiplying by $1/x^2$ can be rewritten $$y^2-2y+1=(y-1)^2=0$$ So $y=1$, or $x-1/x=1$ or $x^2-x-1=0$. Solving this quadratic gives $x=\frac{1\pm\sqrt5}2$, and it is easy to check that $x=0$ doesn't satisfy the original equation, so we have found all solutions.

Parcly Taxel
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  • Thank you for your help. May I ask how you determined that the key was to square $Y$? Also, I don't understand what $x = 0$ has to do with anything. – Novice Sep 12 '20 at 02:50
  • @Novice the original polynomial is palindromic, which means that after making the indicates substitution it can be rewritte entirely in terms of $y$. We need to consider $x=0$ separately because, well, $x=0$ does not define a corresponding $y$ value because divide-by-zero. – Parcly Taxel Sep 12 '20 at 02:52
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$Y^2= X^2-2+\dfrac{1}{X^2}$. So $X^2-1+\dfrac{1}{X^2} = Y^2+1$. Then

$$X^2-2X-1+\dfrac{2}{X}+1/{X^2} = Y^2+1-\left(2X-\dfrac{2}{X}\right)=Y^2-2Y+1$$

  • I think you mean $+1$ instead of $-1$ at the end there, right? – Novice Sep 12 '20 at 02:46
  • Oh yes. Thank you. –  Sep 12 '20 at 02:47
  • Thank you for your help. May I ask how you determined that the key was to square $Y$? – Novice Sep 12 '20 at 02:50
  • @Novice you're welcome. If you notice, the degrees of $x$ in $y$ are $1,0$ and -1 and your expression have 2,1,0,-1,-2. So squaring $y$ will give a function in terms of $x$ with those degrees. –  Sep 12 '20 at 02:53
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Without this hint we can see immediately that: $$x^4-2x^3-x^2+2x+1=(x^2-x-1)^2.$$ Can you end it now?

Michael Rozenberg
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