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Here is another method I wrote where you compute in the Q[√5] field. It can be modified to work over Z[√5] as well:

With yours, both make interesting comparison of how similar problem is solved in different languages.

Why did you define both negate and (-)?

I don't know. I probably shouldn't have. I accept pull requests.

In Haskell you can use Data.Real.Constructible (http://hackage.haskell.org/package/constructible) to compute exactly with arbitrary square roots.