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I know that you can take any element from $R = \mathbf{Z}[\sqrt{-5}]$ like, say, $a = 1 + \sqrt{-5}$, and it will be a principal ideal, but when I used the norms I got this:

$N(a) = N(1 + \sqrt{-5})N(1) = 6,$ so $N(a)$ must be either 1, 2 or 3. $N(a)$ can't be 1 because then $(a) = (1+\sqrt{-5})$ wouldn't be a proper ideal anymore. All that's left is to find out if the equations $a^2 + 5b^2 = 2$ and $a^2 + 5b^2 = 3$ have any solutions. If either one of these has solutions then it confirms that $(1 + \sqrt{-5})$ is a principal ideal, so my question is, are there any?

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    I'm totally sorry if it looks confusing. My bad.$I$do know what we use norms to prove by contradiction that an ideal isn't principal. For example, (3, 1 + sqrt(5)) is not principal becuase there are no solutions to a^2 + 5b^2 = 3. I guess it's not applicable in this situation. Again, soooo sorry people, I am learning all of this stuff for the first time2011-09-18

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Hint: If $b \neq 0$ then $a^2 + 5b^2 > 5$.

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    @crazystudent As some of the others mentioned above, $(1 + \sqrt{-5})$ is principal by its very definition. What you are actually attempting to prove is that it is prime (i.e. it has no proper ideals dividing it). By the hint above you can conclude that if some principal ideal $(a+b \sqrt{-5})$ divides it, you must have $b = 0$. It should be very clear from there.2011-09-18