When is $(-1+\sqrt{5})^x$ a rational number?

For a more elementary answer, note that the reciprocal of $(-1+\sqrt5)^x$, which is $(1+\sqrt5)^x4^{-x}$, must also be rational, hence so is $(1+\sqrt5)^x$. Now you can apply the binomial theorem without worrying about cancellation to see that among the nonnegative integers only $0$ produces a rational number, and negative values of $x$ yield the reciprocals which therefore must also be irrational.


Here is a different approach to the problem. Let $\varphi:\mathbb{Q}(\sqrt{5})\to\mathbb{Q}(\sqrt{5})$ be the automorphism that sends $\sqrt{5}$ to $-\sqrt{5}$. Note that for $a\in \mathbb{Q}(\sqrt{5})$, $a\in\mathbb{Q}$ iff $\varphi(a)=a$. So if $(-1+\sqrt{5})^x$ is rational, then $$(-1+\sqrt{5})^x=\varphi((-1+\sqrt{5})^x)=(-1-\sqrt{5})^x.$$ This clearly can only happen for $x=0$ (for instance, since $\left|-1+\sqrt{5}\right|<\left|-1-\sqrt{5}\right|$ so $\left|(-1+\sqrt{5})^x\right|<\left|(-1-\sqrt{5})^x\right|$ for $x>0$ and $\left|(-1+\sqrt{5})^x\right|>\left|(-1-\sqrt{5})^x\right|$ for $x<0$).

Tags:

Number Theory

Abstract Algebra

Algebra Precalculus

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