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I want to find a cubic such that it meets the following criteria:

  • Has the golden ratio as its only real root
  • Has integral coefficients
  • Has a leading coefficient of $1$ and a final coefficient of $-1$ (this means the imaginary roots have an absolute value of $\frac{1}{\sqrt{\phi}}$)

Does such a cubic exist?

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    $\begingroup$ Where does this problem come from? $\endgroup$ – lhf Jun 18 '15 at 3:45
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If the coefficients are rational and $\dfrac{1+\sqrt 5} 2$ is a root, then $\dfrac{1-\sqrt 5} 2$ is a root.

To see this, suppose you substitute $\dfrac{1+\sqrt 5} 2$ for $x$ and get $0$. What would then happen if you substitute $\dfrac{1-\sqrt 5} 2$ for $x$? When you expand $x^2$ and $x^3$, then wherever $\sqrt 5$ appears, $-\sqrt 5$ would appear, and vice-versa. You won't get $0$ unless the coefficient of $\sqrt 5$ in the total ends up being $0$. If you interchange $\pm\sqrt 5$ then instead of $0$ you get $-0$, but $-0$ is $0$.

The fact that $\sqrt 5$ is irrational is essential here. Suppose $\sqrt 5$ were the rational number $38/17$. Then $17x-38$ would be a polynomial with integer coefficients having $\sqrt 5$ as a root. The argument in the paragraph above assumes the radical cannot vanish like that.

This is very much like the proof that if the coefficient are real and $a+bi$ is a root, where $a$ and $b$ are real, then $a-bi$ is also a root.

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No such cubic exists.

The golden ratio is a root of the quadratic $q(x)=x^2-x-1$.

If the golden ratio is a root of a cubic $p(x)$, then it must be the root of the remainder of $p(x)$ when divided by $q(x)$. That remainder has integer coefficients and is either zero or has degree $1$. But the golden ratio is not a root of polynomial of degree $1$ with integer coefficients because it is irrational. Thus the remainder is zero and the cubic must have at least two real roots: those of $q(x)$.

Note that the third condition in the question is not used in this argument.

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  • $\begingroup$ Note that the quotient of the division of $p$ by $q$ also has integer coefficients, so you can conclude that not only does your polynomial have two irrational real roots, its remaining (third) root is actually rational $\endgroup$ – Marc van Leeuwen Jun 19 '15 at 14:11
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Given enough conditions, you can find the cubic.

The cubic satisfying your conditions (not the second one yet) is $$(x-\phi)\left(x-\frac{1}{\sqrt{\phi}}i\right)\left(x+\frac{1}{\sqrt{\phi}}i\right)=0$$ Expanding it gives, $$x^3-\phi x^2 + \frac{1}{\phi}x-1=0$$ Obviously the coefficients are not integral. Therefore, such a cubic does not exist.

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  • $\begingroup$ You forgot an extra $i$ :P $\endgroup$ – Braindead Jun 18 '15 at 3:57
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    $\begingroup$ @Braindead An $i$ for an $i$ and soon the whole world is blind. $\endgroup$ – corsiKa Jun 18 '15 at 17:09
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Hint: Golden ratio satisfies $p^2=p+1, p^3=p^2+p=2p+1$. So the cubic must have $(2p+1)+a(p+1)+bp-1=0$ which implies $p$ is rational.

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It does not exist.

If it does, let it be $f=(x-\phi)(x^2+ax+b)$

Since $(-\phi)b=-1$, so $f=(x-\phi)(x^2+ax+\frac{1}{\phi})=x^3+(a-\phi)x^2+(\frac{1}{\phi}-a\phi)x-1$

define $n=a-\phi$, so $a=n+\phi$, so $\frac{1}{\phi}-a\phi=\frac{1}{\phi}-(n+\phi)\phi=\frac{\sqrt{5}+1}{2}-n\frac{\sqrt{5}-1}{2}-(\frac{\sqrt{5}-1}{2})^2=(\sqrt{5}-1)(1-\frac{n}{2})$, is only integer when $n=2$, so $\space f=x^3+2x^2-1$.

However, $x^3+2x^2-1$ has 3 real roots. So such $f$ does not exist.

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