150 III More on Divisibility
Suppose next thatp|d.Thend=pefor somee∈Zandpe,sincedis square-
free. IfP=(p,
√
d),thenP^2 =p(p,√
d,e)=pOd,since(p,e)=1. Since we cannot be in cases (i) or (ii) of Proposition 21, we must be
in case (iii).
Suppose conversely that either (i) or (iii)of Proposition 21 holds. Then the corre-
sponding prime idealPcontainsp. Chooseβ=aandγ=b+cωas in Proposition 14,
so that
P={mβ+nγ:m,n∈Z}.In the present case we must havea =p,sincep∈ Pand 1∈/ P.Wemustalso
havec=1, sincePP′=pOdimpliesac=p. With these values ofaandcthe final
condition of Proposition 14 takes the form
b^2 ≡dmodp ifd≡2or3mod4,
b(b− 1 )≡(d− 1 )/4modp ifd≡1mod4.Thus in the latter case( 2 b− 1 )^2 ≡dmodp. In either case ifpd,then(d/p)=1.
This proves that ifpdand(d/p)=−1, then we must be in case (ii) of Proposi-
tion 21.
Proposition 23Let p= 2. Then, in the statement of Proposition 21,(i),(ii),or(iii)
holds according as
d≡1mod8,d≡5mod8,or d≡ 2 ,3mod4.Proof Since the proof is similar to that of the previous proposition, we will omit
some of the detail. Suppose first thatd≡1 mod 8. IfP=( 2 ,( 1 −
√
d)/ 2 ),thenP′=
( 2 ,( 1 +
√
d)/ 2 )andPP′= 2 ( 2 ,( 1 −√
d)/ 2 ,( 1 +√
d)/ 2 ,( 1 −d)/ 8 ).It follows thatPP′= 2 OdandP=P′.
Suppose next thatd≡2 mod 4. Thend= 2 e,whereeis odd. IfP=( 2 ,
√
d),thenP^2 = 2 ( 2 ,√
d,e)= 2 Od.Similarly, ifd≡3 mod 4 andP=( 2 , 1 +
√
d),thenP^2 = 2 ( 2 , 1 +√
d,( 1 +d)/ 2 +√
d)= 2 Od.Suppose conversely that either (i) or (iii)of Proposition 21 holds. Then the corre-
sponding prime idealPcontains 2. Chooseβ=aandγ=b+cωas in Proposition 14,
so that
P={mβ+nγ:m,n∈Z}.