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I was able to verify with Wolfram Alpha that $x^6-3x^2+1=0$ has $4$ real solutions.

However, how would you answer this question without using a computer program?

Rodrigo de Azevedo
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Joshua Bonet
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3 Answers3

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Let $z=x^2$, then $p(z):=z^3-3z+1$ is a cubic and has one or three roots.

The extrema satisfy $3z^2-3=0$, hence $z=\pm1$, giving the values $p(-1)=3,p(1)=-1$ and the three roots are real. As $p(0)=1$, two of the roots are positive.

Hence there are two complex solutions in $x$ and four real ones.

Parcly Taxel
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It is enough to show that the cubic polynomial $x^3-3x+1$ has exactly two positive real roots.
This is indeed a quite famous polynomial, related to the construction of the regular 9-gon.
The discriminant of $x^3-3x+1$ is $81>0$, ensuring the existence of three simple real roots.
At least two of these roots are positive, since $f(x)=x^3-3x+1$ changes its sign over $(0,1)$ and over $(1,2)$, too. On the other hand they cannot be all positive, since the sum of the roots is zero by Vieta's theorem. We are done: $x^6-3x^2+1$ has exactly four real roots, two of them being positive, the other two being negative (no wonder, since $x^6-3x^2+1$ is an even function).

Jack D'Aurizio
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  • Wait... Did you just say discriminant for cubic? I didn't know that cubic have discriminants too earlier! I used to check existence of real root by taking derivative and checking whether $f(\alpha) \cdot f(\beta)<0$ (usual notations) or not. Is your method easier and shorter than this? – Jaideep Khare Mar 07 '18 at 22:14
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    @JaideepKhare: since we have a reduced cubic, to compute the sign of the discriminant is faster than computing the location of the stationary points. Which is perfectly viable, anyway. – Jack D'Aurizio Mar 07 '18 at 22:14
  • I managed today to factor the three-variable cubic of interest in https://math.stackexchange.com/questions/2678486/how-to-show-that-if-x-y-z-are-rational-numbers-satisfying-x-y-z3-9/2678675#2678675 using the regular 9-gon. I knew of $x^3 - 3x + 1$ through Gauss's method and Reuschle's 1875 book. – Will Jagy Mar 07 '18 at 22:23
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If $f(y)=y^3-3y+1$ then $f(-2)<0, f(0)>0, f(1)<0, f(2)>0.$ So $f(y)=0$ has three real roots, with one negative root. Thus $f(x^2)=0$ has four real roots, two for each positive root of $f.$

Thomas Andrews
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