The euclidean real quadratic fields

MATHEMATICS

THE EUCLIDEAN REAL QUADRATIC FIELDS BY

P. VARNAVIDES (Communicated by Prof. J. F. KoKSMA at the meeting of February 23, 1952)

I. Euclid's Algorithm is said to hold in the algebraic field k((J) if, for any non-zero integers a and {3 of k(()), there is an integer ~ of k(()) such that (I) [N(a- ~{3)[ < [N({3)[.

It is known 1 ) that Euclid's Algorithm holds in the real quadratic field k(VD), where D is a positive square-free rational integer, when (2)

D = 2, 3, 5, 6, 7, II, I3, I7, I9, 2I, 29, 33, 37, 4I, 57 and 73.

In this paper we develop the method of our earlier paper 2 ) to prove the existence of Euclid's Algorithm ink( VD) when D has any of the values {2). 2. Suppose that D is a positive square-free rational integer, greater than I. Write

w=- VD, if D = 2 or 3 (mod 4), w=! (I- VD), if D- l (mod 4). Then the general. integer of k( VD) is of the form x + wy where x

(3)

~ w

t

= VD,

w =!(I+ VD),

and y

are rational integers. Write· ( 4)

f(x, y) = (x + wy) (x + roy) \ xz -Dy2, if D- 2 or 3 (mod 4). = ( xz +xy -l- (D- l) yz, if D =I (mod 4),

so that when x andy are rational integers, f(x, y) is the norm N (x + wy) of the integer x + wy of k( VD). It is clear from (I) that Euclid's Algorithm will hold in k(VD) if for every number ~0 of k( VD) there exists an integer ~ of k( VD) satisfying [N(~

+ ~o)l


So Euclid's Algorithm will certainly hold in k(VD), if for all real x0 , Yo there exist rational integers x, y satisfying (5)

[ f(x + Xo, Y +Yo)[
1) H. CHATLAND and H. DAVENPORT, Canadian Journal of Mathematics, 2, 289-296 (1950) and references given there. It was stated by L. REDEl in Math. Annalen 118, 588-608 ( 1943) that Euclid's Algorithm holds in k ( 97), but this has been shown to be erroneous by E. S. BARNES and H. P. F. SwiNNERTON-DYER (in a paper in course of publication). 2) P. VARNAVIDES, Quart. Journal of Math., 19, 54-58 (1948).

V

112

In the following sections of this paper we prove that if D is any of the integers (2) then for all real x0 , Yo there exist rational integers x, y satisfying (5). 3. Let q;(f) be the lower bound of the numbers M such that for all real x0 , Yo there exist integers x, y satisfying (6)

!f(x

+ Xo, Y +Yo)!<

M.

In a previous paper we proved the following slightly improved form of a theorem of DAVENPORT 2). 1)

Theorem. Let f(x, y) be an indefinite binary quadratic form with discriminant d > 0. Let / 1 be any value of I f(x, y) I which corresponds to coprime integral values of x, y and which satisfies 0 < /1 < d. Write d

a2=-

(7)

I6/i'

and let n be any positive integer. Then ~

"f

~"f

(8)

a 2..--1 "':::: 4'

--- 2 "':::: --- .1 41 "'::::a 2'

if ! (n 2 + 1) ~ a 2 ~! (n 2 + 2n), if !(n2 +2n)~a 2 ~!{(n+1)2+1}.

!

Using this theorem we obtain the results given in the following table, since the discriminant d of f (x, y) is given by d = 4D if D d = D if D

!1

D 2 3 5 6 7 II I3 I7 I9 2I 29 33 37 4I 57 73

P. H. (I946). 1)

VARNAVIDES,

2)

DAVENPORT,

=

11 /1 /1 11 /1

2 or 3 (mod 4), 1 (mod 4).

= I f
= I f (0, I) I = 2 = I f (I, I) I = 2 = If (I, I) I = I = If (3, I) I = 3 = I f (2, I) I = 3

/1 = If (3, I) I = 2 /1 = I f (2, I) I = 3 /1 = I f (2, I) I = 2 !1 = I f (9, 2) I = 5 /1 = If (I, I) I = 3 /1 = If (2, I) I = I !1 = If (5, 2) I = 3 11 = I f (2, I) I = 3 11 = I f (2, I) I = 4 !1 = I /(Io, 3) 1= 4 /1 = 1/(34, 9) 1= 4

I

a2= I:,r

cp (f)

1/8 3JI6 5/I6 I/6 7/36

~ ~

ll/I6 I3/I44 I7/64 I9JIOO 7/48 29JI6 33/I44 37JI44 4I/256 57/256 73/256

~ ~

~ ~

~

I/2 I/2 5/I6 3/4 3/4

I 3/4 ~ I7f32 ~ 5/4 ~ 3/4 ~ I ~ 3/4 ~37 /48 ~ I ~ I ~ 73/64

lac. cit. Proc. Kon. Ned. Akad. v. Wetensch., Amsterdam, 49, 8I5-821

113

When

D

2, 3, 5, 6, 7, 13, 17, 21,33 or 37,

=

we have
Lemma. such that

Suppose fJ is any real number and m is a non-negative integer, (32

(9)

~!

(m2

+ 1).

Let Bm(fJ) =max { !,

(10)

i

(m- 1), fJ2-! (m- 1)2

}.

Then for any real number X 0 there exists a real number X, with X _ X 0 (mod 1), such that

IX2 -

(ll)

J) 2 1

~

Bm(fJ).

Further it is possible to choose X, with X- X0 (mod 1), such that (12)

unless m (b) m (c) m (d) m

(a)

= 0, fJ = 0 and X 0 - i (mod 1), or = 0, {3 2 = ! and X 0 - 0 (mod 1), or ;?: 2, {3 2 = ! { (m- 1) 2 + 1 } and X 0 - im (mod 1), or ;?: 1, {3 2 ;:;, ! (m 2 - 1) and X 0 - - i (m- 1) (mod 1).

Proof. We first prove the result when j32 ~ t· We choose X- X 0 (mod 1) so that 0 ~ I X I ~ !, with strict inequality on the left unless X 0 _ 0 (mod 1) and with strict inequality on the right unless X 0 = i (mod 1). Then

1X2- fJ21

~

!,

with strict inequality unless j3 = 0 and X 0 - i (mod 1) or {J 2 = X 0 - 0 (mod 1). This proves the result when {J 2 ~ t· If j32 > !, and l is a positive integer such that j32 then


(1 2 -! (l- 1)2
Also

!
(l2

+ 1) < !

(m2

{ (l 2 + 1)- (l-

(m- 1) and! (l- I)

!

and

+ 1),

1)2 } =


!l

~

i

(m- 1).

(m- 1).

Thus by (10) Consequently, when {J2

> f,

it is only necessary to prove the result when

1) H. DAVENPORT, "Non-homogeneous Ternary Quadratic Forms", Leml:na 5, Acta Mathematica, 80, 65-95 (1948).

8 Series A

114

m is chosen so that l { (m- 1) 2 + 1 } ~ {1 2 ~ l (m2

{13)

+ 1).

We suppose that m is chosen so that (13) is satisfied and we take X= X 0 (mod 1) so that

! (m-

I X I ~! m; this is possible since the intervals from ! (m-

(14)

1) ~

1) to ! m and from to-! (m- 1) include all values of X (mod 1). Further we have strict inequality on the left of (14) unless

-! m

X0

=! (m-

1) (mod 1),

and strict inequality on the right unless X0

Then, by {14), (J2-lm2

~

!m (mod 1).

-

(J2- x2

~

(J2-l (m- 1)2,

so that using (13), {15)

-! (m-

1) = l { (m- 1)2

+ 1 }-lm2 ~ (J2- X2

~

(32-!(m-1)2,

with strict inequality on the left unless

(J2 = l { (m- 1)2 + 1} and X 0

=! m

(mod 1),

and with strict inequality on the right unless X0

Now

!

(m- 1) (mod 1).

(J2-l (m- 1)2

and Hence, by (10),

!

(m- 1)

>

> l,

max {! (m- 1), {J2 -

l

ifm = 1,

l

if m

>

(m- 1) 2

l.

} =

B ({J).

Consequently, by (15), Further, we have unless either

IX2 - {J2 1 ~ Bm ({J). IX2 - {J2 1< Bm ({J),

+ 1 } and X 0 -! m (mod 1), which case m ;:? 2 as (J 2 > l, or {J2 - l (m- 1) 2 ;:?! (m- 1) and X 0 ! (m- 1) (mod 1), {J 2 = l { (m- 1) 2

in

in which case

(J2 ;:? l (m2- 1).

This completes the proof of the lemma. 5. In the six special cases we have to consider, we apply an integral unimodular transformation to f(x, y). Since f(x, y) assumes the value ± /1 for coprime integers x, y there is an integral unimodular trans-

115 formation which transforms

f (x, y)

into the form

± /1 {
yz}

for some real k. We have to prove that for any real X 1 , Y1 there exist integers u, v such that (16)

l(u+kv+Xo)2-

where

4~i (v+Yo)21< 11~1'

X 0 = X 1 + kY1 , Y 0 = Y1 .

It is clear that in proving this result we may suppose without loss of generality that

(17) Further by consideration of the transformations (u, v)-+ (- u,- v) and (u, v) -+ (1- u,- v) we may suppose that (18)

0 ~ Y0 ~

i

if X 0 = 0 or X 0 =

i·

We begin by discussing the six special cases but we leave the discussion of the case D = 19 to the last as being the most complicated one. Case 1. When D transformation we obtain

11, we take / 1

=

X

f

= 3X

+ 2Y,

(x, y) = - 2 {(X

2. Applying the unimodular

=

y= X

+ Y,

+ % Y) 2

-

l

1

P}.

Now (16) takes the form ( 19) Write (20)

yz ___.-

R2 _ 11 -.~

11

o"'::::nl

f'

<

3

4·

Then (9) is satisfied with m = 2 and, by (10), B 2 ({3) = i· So by the lemma we can satisfy (19) by taking v = 0 and choosing a suitable integer u, unless X 0 - 1 (mod 1), {3 2 = i· (21) Now (17), (18), (20) and (21) imply that

X 0 = 0 and Y 0 = V121 . In this case we take u

=-

1, v

=

2 and we have

I(u + ~ v + Xo)2 - Il (v + Yo)21 = I(- 1 + 5)2 - Il (2 + Vr2r)21 =

1<4) 2 -

( 4.o2

... )2 1 <

i.

Thus in any case there are integers u, v satisfying (19).

116

Case 2. When D = 29, we take / 1 = l. Applying the unimodular transformation X = 2X + Y, y = X + Y, we obtain f(x,y) =-{(X+~ Y)2- 21 Y2}.

Now (16) takes the form (22) Write

l(u

+ ~ v + X 0 ) 2 - 21 (v + Y0 ) 2 1 < {3 2

(23)

_ -

11_9 4

y20 ::::::::: __..-

g~

16

< ~4 <

1.

to

4 •

Then (9) is satisfied with m = 3, and by (10), B 3 ({J) = l. So by the lemma we can satisfy (22) by taking v = 0 and choosing a suitable integer u, unless (24) X 0 - i(mod 1) and fJ2 = i. Now (17), (18), (23) and (24) imply that X0 = In this case we take u

=-

i and Y 0 = V/9 •

1, v = 2 and we have

l(u + ~v +X 0 )2- 2l(v + Y0 )21 = 1(6.5) 2 - 21 (2 + V2"9) 2l = 1(6.5) 2 - (6.503 ... )21 < l.

Thus in any case there are integers u, v satisfying (22). Case 3. When D = 41, we take / 1 = 4. Applying the unimodular transformation X '== 2X + Y, y = X + y we obtain f(x,y) = -4 {(X+ VY) 2 - H Y2}. Now (16) takes the form (25)

I (u

+ \3 v + X 0 ) 2

-

U(v + Y0 ) 2 1 < l·

Write {J2-HY2<1 - 64 0 4

(26)

•

Then (9) is satisfied with m == 0 and, by (10), B 0 ({J) = f· So by the lemma we can satisfy (25) by taking v = 0 and choosing a suitable integer u, unless (27) X0 i (mod 1) and fJ = 0. Now (17), (26) and (27) imply that X 0 = i and Y0 = 0. In this case we take u

I (u

=

2, v

=-

1, and we have

+ \a v + X o)2 - U (v + y o)21

u

= 1(2- 183 + i)2- (-1)21 = IW2 I= i < l· Thus in any case there are jntegers u, v satisfying (25).

u

117

Case 4. When D transformation

= X

we obtain

57, we take / 1

= lOX

+ 3Y,

4. Applying the unimodular

=

+ Y,

y = 3X

f(x,y) = 4 {(X-~ Y) 2

U Y2}.

-

Now (16) takes the form (28) Write

u y~ < l·

fJ 2 =

(29)

Then (9) is satisfied with m = 0 and B 0 ({J) = !· So by the lemma we can satisfy (28) by taking v = 0 and choosing a suitable integer u, unless (30)

X0

= t

(mod 1) and ,8 = 0.

Now (17), (29) and (30) imply that X0

=

t

and Y0

0.

=

In this case we take u = 1, v = 1 and we have

I(u

- ~v

+ Xo)2

= I(l -

=1m

-

2 -

H(v + Yo)21 a+ !) 2 - uI

u1 <

~.

Thus in any case there are integers u, v satisfying (28). Case 5. When D = 73, we take / 1 = 4. Applying the unimodular transformation X = 34X + 15Y, y = 9X + 4Y, we, obtain f(x, y) = 4 {(X - i Y )2 - U Y2} . Now (16} takes the form (31} We have to prove that for all real X 0 , Y0 satisfying (17) there are integers

u, v satisfying (31). We consider a number of cases separately, namely the following: I.

II. III.

(a) (b)

0 ~ Xo ~ (a)

(b)

IV.

0 ~ Xo < X0 =

(a)

(b)

l' l Yo = L~

0.

L 0 < I Yo I < v4__. 73

ll8

Case I. In (a), (31) is satisfiedwith u = v = 0. In (b) we take u v = 1 so that l(u- I v + Xo)2- U(v + Yo)21 = I(- 1 - I + ~) 2 - UI = IU - UI < 1,

=-

l,

and t
Case II. We have so that (31) is satisfied with u = v = 0. Case III. ~ 1(1-

In case (a), (31) is satisfied if we take u = v = l. Indeed

S+ I36) 2 - il (1- V21= l28s\

f 1(1- s+ 0)

-lil61 <

L

ii(l-· ~;/I= 1(0.375) 2 - (0.568 ... )21<}.

2 -

In case (b) we have

- 1= b36) 2 -

uW < x~ - uY5 < L 2

so that (31) is satisfied with u = v = 0. Case IV.

In (a), (31) is satisfied if we take u = v = - l. Indeed

I(- 1 + I + o)2 - U(- 1 + V21 = Im2 -/,61 < i, ~ 1(- 1 + s+ .105)2 -- u(- 1 + ~)21 =I (0.27)2- (o.568o ... )2l ~

3

In (b), (31) is satisfied if we take u

i I(- 3 + s + 0.1<)5) 10

. ~ 1(- 3

2, v

=

~ 1(2 _ ~ + 136)2 _

=

3 and v

=-

1.

2. Indeed

=I(- 1.645) 2 - il -II< L

H (- 2 + ~) 2 1

+ \0 + 0.15)2- ii(-2 + ~)21

·In (c) we take u

( 1(2 _

2 -

=-

<

= 1(-1.6)2- (1.63 ... )21 < }.

1, when (31) is satisfied. Indeed

u(1 + ~)_21 = 1(1.5625)2- (1.568 ... )21 < 1,

s+ .15)2- n(1 + !)21 = I(L525)2 _ 2.56 ... 1 < 1.

Now case (d) is similar to the case (b) of III, and this concludes the proof that k( V73) is a Euclidean field. Case 6. When D = 19 we take / 1 = 5. Applying the unimodular transformation X= 9X + 4Y, y = 2X + Y, we obtain f(x, y) = 5{(X- i Y) 2 - i~ P}. Now (16) takes the form I (u

-

i v + X0)2

-

~~ (v

+

Y0 ) 2 1 < ~,

or (32)

J(5u- 2v

+ 5X0 ) 2 -

19 (v

+ Y0 ) 2J <

5.

119

By taking u except when

V5 =

=

v = 0 in (32) we see that this inequality is satisfied

2.236 ... ~ 5X0 ~ 2.5, 0 ~ Y0 ~~Vi\= .2564 ....

To prove that (32) is also satisfied for integral values of u and v when X 0 and Y0 lie in the above intervals we consider the following cases: Case I (a)

V5 ~ 5X0 ~ 2.5 ,

(b)

~

r5 ~ 5X0 < 5X0

(c)

(d)

J/5 ~ 5X0

In (a) take u ~

l

-

2.5,

y 0-- 5V.!.!01_ l 19

2.5 ,

y 0-- 5Vl.ol_ l 19

V

l.Ol ' } 5 19 -

~ 2.5,

<

y _.-

0.1527 ... ,

= ' '

Jtfs

0 '::::::: 2V f9 .

l, v = l. Then

=-

H- 5-2+ V5) 2 -

~ 1(- 5 - 2 + 2.5)

=

0 ~ Y0 < 5 V 1 i~ 1

2 -

19 (1) 2 1 =~4.764 ... )2 - 191 = 122.68 ... - 19 < 5, 19 (5 1 1 ) 21 < 12o.25- 25.251 ""5.

Vi:

In (b) take u = - I, v = I, when we obtain the same results as in case (a). In (c) take u = 6, v = 5. Then 1(30 -

l

10 + 2. 5) 2 -

In (d) take u

=

1(5- 2 + 2.5) 2

1(5-2+

19(5.1527 ... )2 1

=

1506.25 -

504.45. . .

I<

5.

v = l. Then -

19 (5 ~)21 < 130.25-25.251

V5> -19 (I+ H!f;>21 =

=

5,

1(5.236 ... )2- 19 (1.256 ... )21 k =

·

127.41 ... -29.99 ... 1 < 5.

So case I completes the proof for positive Y 0 • When Y0 is negative the proof is not so simple, as we shall see below. Case II

J/5 ~ 5X0 ~ 2.338, In (a) take u

=

0, v

= -

l

(a)

- 0.09 ~ Y 0 ~ 0,

(b)

- 0.1~7 ~ Y0 ~

(c)

- ~ V159 ~ Y 0 ~

0.09,

-

0.157.

-

l. Then

19 ( -l)J == Jl8.8l. .. - l9J < 5, 2 ~ J(2 + 2.236 ... ) - 19 (- 1.09)21 = 117.94 ... - 22.57 .. ~ 1(2 + 2.338) 2 -

In (b) take u ~ 1(10- 4

=

5.

v = 2. Then

+ 2.338} 2 -19 (2- .157) 21 =

( 1(10- 4 + 2.236 ... }2 -

169.52-64.531

' 1(2.338)2 - 19 (- .157) 2 1 = 15.466 ... - 0.468 ... 1 < 5, 21 = 15 - 1.251 < 5. (2.236 .•• )2 - Ill (- I

Vi\)

<

5,

19 (2- .09) 2 1 = J67.8 ... - 69.3 ... J < 5.

In (c) take u = v = 0. Then

ll

·I <

120

Case III

2.338

In (a) take u

=

~ 1(2 + 2.347} 2 -

? 1(2

+

5X 0

0, v

~

=-

2.347,

- .11

~

(b)

- .15

~ Y0 ~

(c)

- .159~ Y 0 ~ -.15,

19 (-

=

2, v

10 + 2.338) 2

-.165~Y0 ~-.159,

-~V/;~Y0 ~-.165.

191

<

5,

·I <

118.81 ... - 23.4o ..

5.

5. Then

-

=

498.07 ... 1 < 5,

1499.38 ... -

503.92 ... 1 < 5.

19 (- 5.15) 2 1 = 1498.98 ... -

-

-.11,

(e)

~ 1(10 + 10 + 2.34.7)2- 19 (- 5 ..11)21

?!(10 +

0,

~

1. Then

1.11} 2 1 =

=

Y0

(d)

19 (- 1) 2 1 = 118.89 ... -

2.338} 2 -

In (b) take u

~

(a)

In (c) take u = - 6, v = 12. Then ~

1(-

30~

19(12- .159} 2 1 = 12668.962 ... - 2663.976 ... i <5,

24 + 2.338) 2 -

( 1(- 30-24 + 2.347} 2 -19 (12- .159} 2 1 = 12668.03 ... - 2668.02 ..

·I< 5.

In (d) take u = 35, v = 28. Then ~

1(175- 56+ 2.347) 2 - 19 (28- .165) 2 1= 114725.09 ... - 14720.95 ... 1 < 5,

? 1(175- 56+ 2.338} 2 -19 (28- .59} 2 1= 114722.91. .. -14727.30 ... 1<5. In (e) take u = v = 0. ~ 1(2.347) 2

ll (2.338)

-

2 -

The~

19 (- .165) 2 1 = j5.5o8 ... - .517 ... 1 < 5, 19 (- ~

j/Jg) 2 1 =

j5.46 ... - 1.251 < 5.

Case IV

2.347

In (a) take u ~ 1(2 + 2.36} 2

?1(2 +

__:_

2.347} 2 -

In (b) take u ~

=

=

~

0, v

~

=-

- .12 ~ Y 0 ~ 0,

(b)

- .15 ~ Y 0 ~

(c) . (d)

- .16

- .1626

(e)

- .20 ~ Y 0 ~- .1626,

(f)

-~VI;~ Y0 ~

2.36,

191

19 (- 1.12) 2 1 = 118.89 ... -

v

=

2.347} 2 -

~

Y0 ~

-

.12,

~ -

.15,

Y0

~ -

-

.16,

.2.

1. Then

19 (- 1} 2 1 = 119.00 ... -

1(10- 4 + 2.36} 2 -

?1(10- 4 +

5X0

(a)

<

5,

23.83 ..

·I <

5.

2. Then 19 (2- .15} 2 1 = 169.88 ... - 65.02 .... 1 < 5, 19 (2- .12} 2 1 = 169.55 ... -

67.15 ... 1 < 5.

121

In (c) take u

=-

6, v

~ 1( -- 3o - 2.4 + 2.347) 2

= -

( 1(- 30-24 + 2.36) 2 -

12. Then

19 ( 12 - 16) 2 1 = l266~l.03 ... - 2663.52 ... 1 < 5, 19 (12- .15)2 1 = 12666.68 ... - 2668.02 .. -I< 5.

In (d) take u = 35, v = 28. Then ~

1(175- 56+ 2.36) 2 -19(28- .1626)21 = 114728.24 ... -14723.49 .. ·I< 5,

( 1(175- 56+ 2.347) 2 -19(28- .16) 2 1= J14725.o9 ... - 14726.24 ... 1< 5.

In (e) take u = 7, v = 6. Then ~

1(35- 12 + 2.36) 2

?1(35- 12 +

2.347) 2

In (f) take u

=

~ 1(2.36) 2

?J(2.347)

2 -

v

19 (6- .20) 2 1 = 1643.12 ... -

-

639.16 .. ·I < 5,

19 (6- .1626) 2 1 = 1642.47 ... -

-

647.42 .. ·I < 5.

0. Then

=

19 (- .2o) 2 J = 15.56 ... - . 761 < 5, 19 (-

i

Vl

9) 2

1= l5.5o ... -1.251 < 5.

Case V

2.36

~

5X0

~

2.425,

(a)

- .12

~

Y0

~

(b)

- .15

~

Y0

~ -

(c)

- .162 ~ Y 0 ~

(d)

- .188 ~ Y 0

(e)

In (a) take u ~

=

0, v

=-

-! Vt; ~

0, .12, -

~ -

Y0 ~

-

.15, .162, .188.

l. Then

1(2 + 2.425) 2 -19 (-1)21 = 119.58 ... -191 < 5,

( 1(2 + 2.36) 2 -

19 (- 1.12) 2 1 = JI9.oo ... - 23.83 ... 1 < 5.

In (b) take u = 2, v = - 5, when ~! (10 + 10 + 2.425r _ 19 (- 5.12tl_= 1502.88 ... _ 498.07 .. ·I < 5,

( I (10 + 10 + 2.36) - 19 (-5. 15) I - 1501.01. .. - 503.92 .. ·I < 5. In (c) we subdivide the interval for 5X0 into two intervals, namely (c 1 )

2.36 ~ 5X0 ~ 2.39, ~ - .162 2.39 ~ 5X0 ~ 2.425,

(c 2 )

~

Y0

~ -

.15.

In (c1 ) take u = - 6, v = 12. Then ~

1(- 30-24 + 2.36) 2 -19 (12- .162) 2 1 = 12666.68 ... - 2662.62 .. ·I< 5,

( 1(- 30-24 + 2.39) 2 -

In (c2 ) take u

=

2, v

~ 1(10 + 10 + 2.425) 2

( 1(10 + 10 +

2.39) 2 -

19 (12- .15) 2 1 = 12663.59 ... - 2668.02 .. ·I< 5.

=-

5. Then

19 (- 5.15) 2 1 = 1502.88 ... 19 (-

5.162) 2 1 =

1501.31. .. -

503.92 .. ·I< 5, 506.27 .. ·I< 5.

122

In (d) take u

=

7, v

=

6. Then

J(35- 12 + 2.425) 2 -19 (6- .188) 2 1 = J646.43 ... - 64l.80 ... J < 5, ~ 1 (35- l2 + 2.36) 2 - 19 (6 :- .162)2 1 = J643.12 ... - 647.56 ... 1 < 5. ~

In (e) take u

=-

2, v

=-

l. Then

19 (- 1.188) 2 1 = 131.809 ... - 26.815 .. ·I < 5, 19 (- 1.256 ... )2 1 = 131.08 ... - 29.97 ... 1 < 5.

~ J(- 10 + 2 + 2.36) 2 -

( 1<- 10 + 2 + 2.425) 2 -

Case VI 2.425

In (a) take u

=

~

0, v

5X0

= -

~

(a)

- .137

~

(b) (c)

-.170

~

Y 0 ~ 0, Y 0 ~- .137,

- .172

~

Y0

(d)

-l~~ Y0 ~- .172.

2.5,

~-

.170,

l. Then

1(2 + 2.5) 2 - 19 (- 1) 2 1 = 120.25- 191 < 5, ~ 1(2 + 2.425) 2 - 19 (- 1.137)2 1 = 119.58 ... - 24.56 ... 1 < 5. ~

In (b) take u=2, V=-5. Then 1(10 + 10 + 2.5) 2 - 19 (- 5.137) 2 1 = 1506.25- 501.38 .. ·I < 5, {1(10 + 10 + 2.425) 2 - 19 (- 5.170) 2 1 = 1502.88 ... - 507.84 ... 1 < 5. ~

In (c) take u

=

7, v

=

6. Then

1(+ 35- 12 + 2.5) 2 - 19 (6- .172) 2 1 = 1650.25- 645.34 .. ·I < 5, ~ 1(35- 12 + 2.425) 2 - 19 (6- .17) 2 1 = 1646.43 ... - 645.78 ... 1 < 5. ~

In (d) take u

=-

2, v = - l. Then

~ 1<- 10 + 2 + 2.425) 2 -

~ 1(- 10 + 2 + 2.5) 2 -

19 <- 1.172) 2 1 = I3Lo8o ... - 26.098 .. ·I< 5, 19 (- 1.256 ... )2 1 = 130.25- 29.97 ... 1 < 5.

This completes the proof that the field k( V19) is Euclidean.

Khartoum University College, Sudan