This was a result that Dedekind figured out in the 1850s. I've done the calculations for P^2 and found out the real part is indeed <2> but this entirely avoids the imaginary part.
He also said that PQ = <1 + sqrt(5)*i>
He also said that PQ = <1 + sqrt(5)*i>
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Since P = <2, 1 + i√5> and Q = <3, 1 + i√5>:
P^2 = <2^2, 2(1 + i√5), (1 + i√5)^2>, by multiplying all possible pairs of generators .
......= <4, 2 + 2i√5, -4 + 2i√5>
......= <4, 2 + 2i√5, 2i√5>, by adding first and last generators
......= <4, 2, 2i√5>, by subtracting second and last generators
......= <2>, since first anf third generators are Z[i√5]-multiples of 2.
PQ = <2 * 3, 2(1 + i√5), (1 + i√5) * 3, (1 + i√5)^2>, by multiplying all possible pairs of generators
......= <6, 2(1 + i√5), 1 + i√5, (1 + i√5)^2>, by subtracting second and third generators
......= <(1 + i√5)(1 - i√5), 2(1 + i√5), 1 + i√5, (1 + i√5)^2>
......= <1 + i√5>, since all other generators are Z[i√5]-multiples of 1 + i√5.
I hope this helps!
P^2 = <2^2, 2(1 + i√5), (1 + i√5)^2>, by multiplying all possible pairs of generators .
......= <4, 2 + 2i√5, -4 + 2i√5>
......= <4, 2 + 2i√5, 2i√5>, by adding first and last generators
......= <4, 2, 2i√5>, by subtracting second and last generators
......= <2>, since first anf third generators are Z[i√5]-multiples of 2.
PQ = <2 * 3, 2(1 + i√5), (1 + i√5) * 3, (1 + i√5)^2>, by multiplying all possible pairs of generators
......= <6, 2(1 + i√5), 1 + i√5, (1 + i√5)^2>, by subtracting second and third generators
......= <(1 + i√5)(1 - i√5), 2(1 + i√5), 1 + i√5, (1 + i√5)^2>
......= <1 + i√5>, since all other generators are Z[i√5]-multiples of 1 + i√5.
I hope this helps!