We shall assume that the coefficients of (i) are in 
tegers, and in that case call the root q a quadratic integer. 
Then 2a and 4(a 2 —db 2 ) are integers. Thus 4dh 2 
is an integer. But d is an integer not divisible by a 
perfect square > 1. Hence 4b 2 has unity as its denomina 
tor, so that it and 2b are integers. Thus a = ^a, ¿ = ^/5, 
where a and /5 are integers. Since a 2 — db 2 shall be an 
integer, a 2 —df3 2 must be a multiple of 4. 
If d is even, a 2 must be even and hence a multiple of 
4. Thus also d/5 2 must be a multiple of 4. But d is 
not divisible by the square 4. Hence /5 2 is even. Thus 
a and /5 are both even. Hence, if d is even, q is a quad 
ratic integer if and only if a and b are both integers. 
If d is of the form 4&+3, then a 2 — d/5 2 and hence 
also a 2 + /5 2 must have the remainder zero on division 
by 4. According as an integer is even or odd, its square 
has the remainder o or 1. Hence a and /5 are both even. 
If d is of the form 4&+1, then a 2 —d/5 2 , and hence 
also a 2 —/5 2 , must have the remainder zero on division 
by 4, so that a and /3 are both even or both odd. Hence 
q = a+bVd is now a quadratic integer if and only if 
a and b are both integers or both halves of odd integers. 
These two cases may be combined by expressing q in 
terms of the quadratic integer 6 defined by 
0 —^(r + l 7 d), d — 4&+1, 
instead of in terms of Vd itself. First, if a and b are 
integers, then x = a — b and y — 2b are integers and 
q = x+yd. Second, if a = ^{2r-\-i) and ¿=^(25+1) 
are halves of odd integers, then x = r — s and y = 2S+1 are 
integers and q = x-\-yd. 
QUADRATIC NUMBERS