# Which numbers have square roots in $\mathbb Q(\phi)$?

Take the field $Q(\phi)$ when $\phi$ is the golden ratio. Some elements do have square roots in $Q(\phi)$: $$\sqrt{1+\phi} = \phi$$ since $\phi^2=1+\phi$. Also, $(1+2 \phi)^2=5+8 \phi$ so $$\sqrt{5+8 \phi}=1+2 \phi.$$

While the diagonal of a golden rectangle ($1 \times \phi$) is of length $\sqrt{1+\phi^2}=\sqrt{2+\phi}$. What is the square root of $2+\phi$?

Which elements in $Q(\phi)$ have square roots in $Q(\phi)$?

• Use \times insted of x to denote multiplication: 1 \times \phi renders as $1 \times \phi$. Sep 21, 2015 at 13:03
• The title does not really reflect the question.
– lhf
Sep 21, 2015 at 13:08
• All algebraic numbers have square roots... And these are algebraic numbers. In other words, the set of algebraic numbers is closed under square roots (and cube roots, and all roots).
– lhf
Sep 21, 2015 at 13:09
• You can give an answer closely analogous to the answer to the corresponding question for $\mathbb{Q}$ except you need to be more careful about units. The point is that $\mathbb{Z}[\phi]$ has unique factorization, so elements of $\mathbb{Q}(\phi)$ also have unique factorizations (with integer exponents), up to units. A necessary condition is that all of these exponents are even, and then you also need to know which units are squares on top of that. Sep 21, 2015 at 22:31
• It might help to remember that $$\phi = \frac{1 + \sqrt{5}}{2}.$$ Then $$(a + b \phi)^2 = \left(a + b\left(\frac{1 + \sqrt{5}}{2}\right)\right)^2 = a^2 + ab + \frac{3b^2}{2} + ab\sqrt{5} + \frac{b^2 \sqrt{5}}{2}.$$ Or it might not, I'm not sure where I was going with this.
– Lisa
Sep 24, 2015 at 21:15

$$\mathbb Q(\phi)$$ is the linear span (over the rationals $$\mathbb Q$$) of $$1$$ and $$\phi$$. $$(x + y \phi)^2 = x^2 + 2 x y \phi + y^2 (1+\phi) = (x^2 + y^2) + (2x+y) y \phi$$ Thus for $$a,b \in \mathbb Q$$, $$a + b \phi$$ has a square root in $$\mathbb Q(\phi)$$ iff there is a solution in the rationals of \eqalign{a &= x^2 + y^2\cr b &= (2x+y) y\cr} Eliminating $$y$$, we get $$5 x^4 + (2b - 6 a) x^2 + (a-b)^2 = 0$$ so what we need is that this polynomial has a rational root $$x$$. The Rational Root Theorem may be useful (Edit :fixed link typo). Also, it must have real solutions, so the discriminant $$16 (a^2 + a b - b^2) \ge 0$$.
$\sqrt{2+\phi}$ is a solution of irreducible polynomial $X^4-5X^2+5$, so it is not in $\mathbb Q(\phi)$.