# Is it true that $(4+\sqrt{14})(4-\sqrt{14}) = (4+\sqrt{14})^2$ in $\mathbb{Q}(\sqrt{14})$?

Is it true that $(4+\sqrt{14})(4-\sqrt{14}) = (4+\sqrt{14})^2$ in $\mathbb{Q}(\sqrt{14})$? I am going through the solution of a problem I'm working on and this seems to be what they are saying. If its true, why so? I see that the left hand side is $(2)$ and the right hand side is $(2(15+4\sqrt{14}))$ but then what?

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No it is not true. – Calvin Lin Nov 4 '13 at 2:51
It is simply not true. – Don Larynx Nov 4 '13 at 2:52

It's true, if you are talking about ideals.

Clearly, $(2(15+4\sqrt{14})) \subset (2)$. For the other inclusion, note that $(15+4\sqrt{14})(15-4\sqrt{14}) = 1$.

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No. $\mathbb{Q}(\sqrt{14})$ is an integral domain, so if we had equality it would imply that

$$4 - \sqrt{14} = 4 + \sqrt{14}$$

by cancelling a factor of $4 + \sqrt{14}$ from each side. Rearranging would lead to $2 \sqrt{14} = 0$, which is false.

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