Dear student 
For a+b5–√∈Z[5–√] define its norm N by:
N(a+b5–√)=a2−5b2.
Note that:
N((a+b5–√)(c+d5–√))=N((ac+5bd)+(ad+bc)5–√)=(ac+5bd)2−5(ad+bc)2
=a2c2+10abcd+25b2d2−5a2d2−10abcd−5b2c2
=a2c2−5b2c2−5a2d2+25b2d2=(a2−5b2)(c2−5d2)=N(a+b5–√)N(c+d5–√).
So N is multiplicative.
Clearly, if a+b5–√ is a unit, then, we must have N(a+b5–√)=±1.
Suppose, for the sake of argument, N(a+b5–√)=−1, so:
1=5b2−a2. If |a|≥|b|, then:
1=5b2−a2≥4b2. This is a contradiction, unless b=0, in which case we have: 1=−a2, which is impossible.
Similarly, if |a|
1=5b2−a2>4a2 which forces a=0, leading to a similar contradiction.
Thus if a+b5–√ is a unit, it has norm 1. The converse is clear: if N(a+b5–√)=1, then a2−5b2=1, in which case:
a+b5–√ has inverse a−b5–√. For example, 9+45–√ is a unit.
Now if 2=(a+b5–√)(c+d5–√), then:
4=(a2−5b2)(c2−5d2).
As we saw above, NO element of Z[5–√] has norm -1, and if one of the factors of 2 has norm 1, it is a unit, so it doesn't count (a ring element is reducible only if it factors into two non-units).
This means if 2 is reducible, its non-unit factors must each have norm ±2.
The same argument we used to show no element of Z[√-5] has norm -1 also works to show that no element has norm -2.
So we must have a2=5b2+2. Here, we can use a clever trick: