TITLE: Finding cubic with golden ratio as root
QUESTION [11 upvotes]: 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?

REPLY [2 votes]: 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.