Proof that the ideal (2,1+sqrt(-5)) is nonprincipal in Z[sqrt(-5)]

I'm a little stuck here so I was hoping someone could help me on this problem.

My proof by contradiction starts out by assuming that (f(x)) = (2, 1+sqrt(-5)) so N(a + bsqrt(-5)) divides both 4 and 6. Then I have these two equations : a^2 + 5b^2 = 4 and a^2 + 5b^2 = 6. Would I then have to do it by case studies?

Good start.

Suppose that <2, 1 + √(-5)> = for some integers a, b.

So, 2 = (c + d√(-5))(a + b√(-5)), and 1 + √(-5) = (r + s√(-5))(a + b√(-5))
for some integers c, d, r, s.

Taking norms yields
4 = (c^2 + 5d^2)(a^2 + 5b^2), and 6 = (r^2 + 5s^2)(a^2 + 5b^2).

Hence, (a^2 + 5b^2) | 4, and (a^2 + 5b^2) | 6.
==> (a^2 + 5b^2) | 6 - 4 = 2.

So, a^2 + 5b^2 = 1 or 2.

(i) If a^2 + 5b^2 = 1, then a + b√(-5) is a unit in Z[√(-5)].
This contradicts <2, 1 + √(-5)> being a proper ideal of Z[√(-5)] (that is, 1 is not in this ideal).

(ii) If a^2 + 5b^2 = 2, then this has no integer solution (a, b).

Either way, we conclude that <2, 1 + √(-5)> is not principal in Z[√(-5)].

I hope this helps!