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Is there any number $a+b\sqrt{5}$ with $a,b \in \mathbb{Z}$ with norm (defined by $|a^2−5b^2|$) equal 2?

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It is perhaps worth noting that the elements of $\mathbb Q(\sqrt{5})$ which are integral over $\mathbb Z$ include $\frac {1+\sqrt 5} 2$ (integral in this context means numbers which are roots of monic polynomials with integer coefficients). So this is just a check that the question is correctly posed for its context. –  Mark Bennet Jul 1 '12 at 18:51
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up vote 9 down vote accepted

No.

Since $a^2-5b^2\equiv a^2\pmod{5}$, the values of $a^2-5b^2$ must be congruent to either $0$, $1$, or $4$ modulo $5$, since those are the only squares modulo $5$. But neither $2$ nor $-2$ satisfy this condition, so it is impossible to have $a^2-5b^2=2$ or $a^2-5b^2=-2$ with $a,b\in\mathbb{Z}$.

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Hint $\rm\,\ 2\:|\:a^2\!-\!5b^2\! = (a\!-\!b)(a\!+\!b)\!-\!4b^2\Rightarrow\, 2\:|\:a\pm b\:\Rightarrow\:2\:|\:a\mp b\:\Rightarrow\:4\:|\:a^2\!-\!5b^2$

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