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?
Best Answer
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.