Show that a product lies in $(\mathbb{Q}^*)^2$

Question

Consider $\mathbb{Q}(\sqrt d)$, where $d\in\mathbb{Q}$ but $d\notin\mathbb{Q}^2$.

Let $a,b,c$ be distinct rational numbers.

Assume the following facts:

• $(a-\sqrt{d})(b-\sqrt{d})(c-\sqrt{d})\in(\mathbb{Q}(\sqrt d)^*)^2$ (group of nonzero squares)
• $(a+\sqrt{d})(b+\sqrt{d})(c+\sqrt{d})\in(\mathbb{Q}(\sqrt d)^*)^2$
• $(a^2-d)(b-a)(c-a)(a-\sqrt{d})(b-\sqrt{d})(c-\sqrt{d})(a+\sqrt{d})(b+\sqrt{d})(c+\sqrt{d})\in(\mathbb{Q}^*)^2$

My Question:

Is this information enough to conclude that $(a^2-d)(b-a)(c-a)$ also lies in $(\mathbb{Q}^*)^2$? It certainly lies in $(\mathbb{Q}(\sqrt d)^*)^2$.

Any help/hints would be appreciated! :) Thanks

Answer 1

A few hints to help you along the way: consider the 'conjugation' operation $\overline{a+b\sqrt{d}}\mapsto a-b\sqrt{d}$. Then since your conditions imply that there exist $p,q$ such that $(a+\sqrt{d})(b+\sqrt{d})(c+\sqrt{d}) = (p+q\sqrt{d})^2$, you can apply conjugation to both sides of this (why?) to get that $(a-\sqrt{d})(b-\sqrt{d})(c-\sqrt{d})=(p-q\sqrt{d})^2$. Now can you see how to continue from here?