Finding the inverse of $2+\sqrt{5}+2\sqrt{7}$

Question

Finding the inverse of $2+\sqrt{5}+2\sqrt{7}$ in the field $\mathbb{Q}(\sqrt{5},\sqrt{7})$.

I know that all the elements of $\mathbb{Q}(\sqrt{5},\sqrt{7})$ are of the form:

$a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35}$, where $a,b,c,d \in \mathbb{Q} $

So I could simply solve: $(2+\sqrt{5}+2\sqrt{7})(a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35})=1$

Which I believe works. Is there a better way of solving this problem?

This will work - you end up solving four equations in four unknowns. This view of the problem actually leads to the view of $2+\sqrt{5}+2\sqrt{7}$ as a $4\times 4$ matrix. But, as others have pointed out, there is a better way. — Thomas Andrews, Apr 15 '15 at 17:59

score 2 · Accepted Answer · answered Apr 15 '15 at 17:45

2

Hint: Consider the product $(2+\sqrt{5}+2\sqrt{7})(2-\sqrt{5}+2\sqrt{7})(2+\sqrt{5}-2\sqrt{7})(2-\sqrt{5}-2\sqrt{7})$.

answered Apr 15 '15 at 17:45

André Nicolas

507,029

I think I see your idea, that's like a "conjugate" right? – King Ghidorah Apr 15 '15 at 17:46
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Yes, it is essentially "all the conjugates." @KingGhidorah – Thomas Andrews Apr 15 '15 at 17:50
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Yes, the product I mentioned will be rational, indeed an integer, and now we have an expression for the reciprocal. This is not the most efficient way of computing the reciprocal, but it has the advantage that it is readily extended. – André Nicolas Apr 15 '15 at 17:51

egreg · Answer 2 · 2015-04-15T17:57:37.927

2

\begin{align} \frac{1}{2+\sqrt{5}+2\sqrt{7}} &=\frac{2+\sqrt{5}-2\sqrt{7}}{(2+\sqrt{5})^2-28}\\ &=\frac{2+\sqrt{5}-2\sqrt{7}}{-19+4\sqrt{5}}\\ &=\frac{(2+\sqrt{5}-\sqrt{7})(-19-4\sqrt{5})}{361-80}\\ &=\dots \end{align}

edited Apr 15 '15 at 17:57

answered Apr 15 '15 at 17:48

egreg

238,574

Finding the inverse of $2+\sqrt{5}+2\sqrt{7}$

2 Answers2