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Chapter 19 Representation of Numbers by Special Binary Quadratic and Quaternary Quadratic Forms
CHAPTER 19
Representation of Numbers by Special Binary Quadratic and Quaternary Quadratic Forms 1. We deal here with the representation of numbers by special quadratic forms in 2 and 4 variables. Such questions are included in the general theory of the representation of numbers by quadratic forms which is discussed in Chapters 18 and 20. Many results however can be found very simply without using the general theory, and this had been long known when a = b = 1. The method now expounded is due to Thue, who applied it in a few instances, as did also Nagell, but it is capable of wider application.
The equation ax2
+ by2 = m.
We suppose that m > 0 and that (ab, m) = 1. Write f ( x , y ) = ax2 by2 = m,
+
We shall suppose that ( x , y ) = 1 and so ( x y , m) = 1. Solvability of equation (l), if ( x , y ) = 1, implies the existence of an integer k such that ak2 b = 0 (mod m). (2)
+
This imposes conditions on a and b. Thus if m is odd, the Legendre symbol ( - a b / p ) = 1 for every prime p dividing m. We first prove the l s 2 Theorem 1 If the congruence
+b
O(modm) is soloable, integers x , y , not both zero, exist such that ak2
ax2
3
+ by2 = M m
(3)
for some integer M with M < 2dab if a > 0, b > 0, and IMI < d/labl if ab < 0. Put x = k X + m Y,y = X . Then by Minkowski's theorem on linear forms in Chapter 5, there exist integers (A', Y ) # (0,O)such that 1x1 <
fdm,
where 1 is any positive constant.
lyl
< I-wm,
19.
165
BINARY A N D QUATERNARY QUADRATIC FORMS
Hence if a > 0, b > 0. f(x,y ) <
if we take l 2
=
(UP+ b/I2)m < 2 d a b m ,
dbz.
If ah < 0,
1f ( x , y)l if we take l 2
=
Further
f ( ~y ),
< max( la1I2rn,lb1Pm) <
d l q m,
dml.
+ m Y)2+ bX2 = (ak2 + b ) X 2 + 2akmXY + am2 Y 2 = 0 (mod m )
=
a(kX
=
Mm.
The problem now is to determine whether M = 1 is possible. Many of the possibilities for M can be eliminated by congruence considerations. Suppose that m is odd. I. congruences mod 4 or mod 8. Thus all to mod 4, a 3 1, b = -1 excludes M = 2 (mod 4) since then x and y are odd and x2 - y 2 I 0 (mod 4). If a = I , b = 1, 2, then solvability for M = 0 (mod 4) implies solvability for M / 4 since now x = y = 0 (mod 2). 11. congruences mod q, q = 3, 5,7,. . ., a prime. If (-ab/q) = - 1, then M $ 0 (mod q). 111. sometimes it may be shown that if M = M,, say, is possible, then M = M 2 is also possible and conversely. For this we use the results,
if then and if then
+ by2 = m, ax: + by? = ml, (axx, + byy# + ab(xyy,- x , ~ =) mm,; ~ x 2 + aby2 = m, as: + by: = m,, a(xx, + byy,)' + b(sy, - ays,)2 = mm,. ax2
(4)
+
(5)
1V. If (m, ab) = 1, and a is square free, then solvability of ax2 by2 = am implies that of x2 aby2 = m and conversely. We suppose for simplicity that m is an odd prime p since solvability when 171 is composite follows easily from the identities in 111. Let us now consider the detailed results. We begin with the case a > 0, h > 0. We take a = 1 and then
+
.'i2
+ by2 = My,
M < 2db,
and (- b/p) = I is the necessary condition for solvability. We write this in a slightly different form, for example, if b e 1 (mod 4), then (- l / p ) ( p / b )= 1,
166
DIOPHANTINE EQUATIONS
and ifb E 3 (mod 4), then (p/b) = I . l f t b = r = 3 (mod 4), then ( 2 / p ) ( p / r )= 1, but if 46 = r = 1 (mod 4), then ( - 2 / p ) ( p / r ) = 1. 2. It will suffice if we give the results for square free b < 20 and give details for a few values of b. M = I i f b = I , 2,3, 7 and (-b/p) = I Take for example b = 7 and so M < 2/28 = 1 , 2 , 3,4, 5 . We can exclude M = 2 , 4 since then x E y = 1 (mod 2 ) and x2 + 7y2 = 0 (mod 8). We can exclude M = 3, 5 since (-7/3) = - 1 = ( - 7 / 5 ) . M = I i f b = 5providedthat(-l/p) = ( p / 5 ) = l . I f ( - l / p ) = ( p / 5 ) = -1, then M = 2 or 3 and either value implies the other. The condition (- 5 / p ) = 1 gives (-lip) = ( p / 5 ) . Here M < d% = I , 2, 3,4. If M = 4, then x = y = 0 (mod 2) and implies M = 1. For this, ( p / 5 ) = 1 and so (- l/p) = 1. Then M = 2 or 3 can arise only when (2p/5) = 1 or (3p/5) = I , i.e. ( p / 5 ) = - 1. Suppose now x2 5y2 = 2p. Multiply by l 2 5. l 2 = 6. Then from equation (4),
+
+
x2
+ 5Y2 = 3p,
+ 5y = 2 x , 3 s = x + 5Y,
where
.x
and so
x 31'
- y = 2Y, =
x-
Y
If x and y are given, x = y (mod 2 ) and so A', Yare integers. If X and Yare given, we can choose Y so that X = Y (mod 3), and then x, y are integers. M = I ifb = 6 provided that ( 2 / p ) = (p/3) = 1 but if ( 2 / p ) = (p/3) = - 1, then 2 . 2 + 3y2 = p . M = 1 ifb = I0 provided that ( - 2 / p ) = ( p / 5 ) = 1 but if ( - 2 / p ) = ( p / 5 ) = - 1, then 2x2 5y2 = p . M = I or 4 jf h = I 1 or 19, but we cannot remove the ambiguity. The solvability of s2 by2 = Mp means that
+
+
s 2
+ sy +
31'2
=p,
A-2
+ xy + 5y2 = p
are both solvable. M = l i f b = 13providedthat(-l/p)=(p/13)= l b u t i f ( - l / p ) = ( p / 1 3 ) = - 1, then A4 = 2 or 7 and M = 2M = 7. M = I i f b = 15 provided that (p/3) = ( p / 5 ) = 1 but if ( p / 3 ) = ( p / 5 ) = - I , then 5 2 + 3y2 = p . Different kinds of results arise when b = 14, 17. Suppose first that b = 14. Then if (2/p) = (p/7) = 1, either M = 1 or 2 2 + 5y2 = p and we cannot remove the ambiguity. If (2/p) = (p/7) = - I , then either M = 3 o r 2x2 5y2 = 3p, and we cannot remove the ambiguity. When h = 17, M = 1 o r 2 if ( - lip) = (p/17) = 1 and we cannot remove the ambiguity. I f (- I/p) = (p/17) = - 1 , M = 3, 6, 7, each implying the other two.
+
19.
I67
B I N A R Y A N D Q U A T E R N A R Y Q ~ J A D K A T I CFORMS
We now consider b = 1 1 in detail. Here A.i < 2d71 = I , 2 , 3, 4, 5, 6. We can exclude M = 2, 6 since s2 + 1 ly2 5 0 (mod 4). If M = 4, the case when x, y are even reduces to M = 1. We show that the case when s and y are both odd, and the cases M = 3, 5, each imply the other two. Suppose s2 + 1 l y 2 = 4p. From l 2 + 1 I . l 2 = 12, we deduce A2
+ l l B 2 = 3p,
where
4A=.~+Ily,4B=.~-y;
3y=A-B,
3.\-=A+lIB.
By choice of sign of y , A and B are integers. If A , B are given, we can take A = B (mod 3), and then x, y are integers. Suppose next that x2 1 1y2 = 5p. From 22 11. l2 = 15, we deduce
+
C2
+ 110'
+
=
3p,
where 5C
=
+
2 . ~ 1 IJ,
5D
=
s - 2 ~ ; 3.u
=
2C
+ 11D,
3 , ~= C - 2 0 .
We can take x = 2y (mod 5) and later C = - D (mod 3), and then have integer correspondences. We show that if these three cases occur, then M = 1 cannot occur and conversely. This is easy t o see from X 2 XY 3Y2 = p, but the result is a particular case of the
+
Theorem 2 The equations ax2
+ by2 = Ip,
ax:
+ by:
=
+
mp, (.u, y)
cannot both be solrable if a > 0, b > 0, (ah, p ) We may take to mod p, x
E
s1 = k y ,
ky,
Since
(asx,
on putting
axx,
=
=
1, (x,, y , ) = 1.
I, I #
ak2 + b
where
in, ltiz
< ah.
= 0.
+ byy,)2 + ah(.uy, = Imp2, + byy, = p X , xy1 - s l y = p Y ,
where X , Yare integers, then X2
+ ahY2 = In?.
Hence Y = 0, i.e. s / y = and then s = ? sl, y = f y , , and this is not so. 3. We turn now to the case ab < 0. We write equation ( I ) in the form xl/jll,
5 2
-
cy2 = Ep,
E
=
f 1.
168
DIOPHANTINE EQUATIONS
Now Some results may be mentioned. M = I if c = 2, 5, 13, 17. We also have here M = - 1 since x2 - cy2 is solvable for these c on noting y = 5 for c = 13. M = k I w h e r e c = 3 , 6 , 7 , 11,14,19,andif
and if
E
c
= 0 (mod 2),
=
=
M
k 1 or
=
* 2 when
=
(5)
=
=
+I,
=
+I.
-1
(6)
=
sign corresponds to M The M = k I i f c = 15and
(t)
(5)
($) k 1 , ($) *1,
1 (mod 2),
c
=
k 1. 1, p
=
f 1 (mod 4).
c = 10 according as
(f)= (5) = I or - I .
The details are comparatively simple except for c Here
19 and so we give them.
=
x2 - 1 9 ~ '= Mp.
I
=
1. We show that any of
4' - 19.1'
=
-3,
We can exclude IM1 = 4 since this reduces to IM M = 1, - 2, - 3 implies each of the other two. For from
X'
- 19y3 = p ,
A' - 19B2 = -3p,
where A
=
4~ - 19y, B
=
x - 4 ~ ; 3~ = -4A
Since we can take A = B(mod 3), M = 1Next M = - 2 tf M = -3. For from S'
-
1 9 ~ '= -2p,
M
+ 19B, =
3y
=
-A
+ 4B.
-3.
5' - 19.1'
=
6,
C' - 1 9 0 ' ~-3p, where 2C
= 5~
- 19y, 2 0
=
x - 5y;
3~
=
5C - 19D, 3y
=
C - 5D.
19.
BINARY AND QUATERNARY QUADRATIC FORMS
169
The result is now obvious since we can take x = y (mod 2), C = - D (mod 3 ) . Similarly, we can show that each of M = - 1 , 2 , 3 implies each of the others. Hence when (- I/p) = I , we can take M = 1 since M = - 1 is excluded by the congruence x2 - 19y2 = - p = - 1 (mod 4). So when ( - I/p) = - I , we have M = - 1. 4. We now consider the representation of numbers m, which we may suppose
square free, by the special quaternary forms
j
= f(x,
y , z , w) = x2
+ bcy2 + caz2 + abw2.
(6)
These forms admit a composition process. Thus if
+ bcy: + caz: + abw:,
fl = f(s1, y,, z,, w,) = xf then where
ffl = f(-hY 2 , z2, )+>2),
fz
=
XZ
= . Y X ~-
(bcyy,
+ CaZ.7.1 + abrtlw,),
(7)
+ xy, + U(ZM’, - WZ1), = + + b(rt’y, yw!,), = wx1 + + c(yzl zy,).
yz = y s , ~2 11’2
XZ,
ZS,
-
XM’,
-
Hence iff represents both m and m,, thenSrepresents mm,. Theorem 3 Integers (s,y , z ,
it,)
# (0, 0, 0, 0 ) exist such that
j ( x , y , z , w) = Mm,
prolitled that the congruence cA2
I M 1 < d2FFC1,
+ bB2 + a E O(modrt7)
(8) (9)
is solaable.for A and B.
It is shown in Chapter 6 that the congruence is solvable if (m, abc) = 1, = c A X + b B Y + m Z , y = B X - A Y + mW,z = X, = Y. From equations (6) and (9),
m f. O(mod4).Puts 11’
,f(s,y,
2 , I+!) =
F ( X , Y, 2,W ) = Mm,
(10)
where M is an integer, and the determinant D of F is a2b2c2m4. There are several methods for estimating a minimum value p o r lower bound for a quaternary form F ( X , Y , Z, W ) of determinant D. The most elementary is to use Hermite’s* lpl < (4)3’2qrq, and so M < 1 . 5 4 d m l . Next perhaps is the application of Minkowski’s theorem on linear forms in Chapter 5. Thus integers (X,Y , Z , W ) # (0, 0, 0, 0) exist such that, if 6 = q/labcl, then f 61x1 < 2- labanl, Sly1 < SIzI < djhnll, 61wl < dlGl. This gives IMI < 4dlad;l. The estimate (8) follows from the well known 4 more precise results of Korkine and ZolotarefP lpl < 2/4\01. These ~~
~~
,I%[’%
170
DIOPHANTINE EQUATIONS
estimates are usually given for positive definite forms, but they also hold for indefinite forms although the results are not very sharp. We commence with the case c = 1 and so
f = x2 + by2 + az2 + abw2 = Mm, IMI
< dm,
+
(1 1)
and A 2 + bB2 a = 0 (mod m), (m, ab) = 1 (12) The classical case is a = b = 1, Hermite’s result gives M < 1-54, i.e. M = 1. Minkowski’s estimate gives A4 < 4 and so M = I , 2, 3. The case M = 2 can be written as
(F)’ + (y)’ + + + (T) ( Tz ) z
U’
Z-MM”
=
m,
since either x, y , z , M: are all even or odd or only two odd, say x, y. The case M = 3 can be written as
(“ + ;
+;
+ 3”
1’
)Z - 2)Z (x - w)’ (x -y = 171, (y M’ since we may suppose s = 0 (mod 3), y = z E w = 1 (mod 3). The method shows that m is representable by f in equation ( 1 I ) when b = I , a = &3, 5 5 , -7, -11 and also when a = 5 2 , b = 3, -3. It will suffice to take a few of these cases.
+; +
b
=
If M If x
l,a =
=
+
M <
3
2,
s2
+
d6 = 1, 2,
(m,3)
+
=
+ y2 + 3(z2 + w2) = 2m.
1.
(13)
= y (mod 2), then z = w (mod 2), and
and this is the case M = 1. If x = y 1 (mod 2), then z
+
1
= 14’
+ 1 (mod 2), and so from equation (13)
+ 3 = 2m (mod 4),
m
= 0 (mod 2).
Hence M = 1 if m = 1 (mod 2). I f m = 0 (mod 2), the argument applies to t m , and since 2 is representable b y f so is m . Finally if m z 0 (mod 3), +m is representable by f and so is 3 and hence also m. b = I , a = -3 (m,3) = 1. ] M I = 0, 1,2, Since - 1 is representable, we need only consider M 3 0. Clearly M = 0 is impossible since this requires x 2 y = 0 (mod 3) and then z = w = O(mod 3). If M = 2, the argument above applies when x = y (mod 2), z = iti (mod 2). For .Y E y 1 (mod 2), z = it’ 1 (mod 2), we need only consider
+
x
= 1 (mod 2),
y
= 0 (mod 2),
z
+
= I (mod 2),
w
= 0 (mod 2),
19.
171
BINARY AND QUATERNARY QUADRATIC FORMS
Since
12
-
3.12 =
-2,
we have
(T)
( x + 3z 7 - 3 ) +' ( y + 3w 7 -3 y) + w 2 =' -m, (x+)2
and this is the case M = - 1. The case m = 0 (mod 3 ) is dealt with as above. b = I , a = 5, then (m, 5) = 1 and this case is of some interest. Here
x2
+ y 2 + 5(z2 + w2) = Mm,
M
1,2, 3.
=
Clearly M # 1 for all m since 3 is not representable. It can be shown from the theory of the ternary quadratic that all integers > 3 are representable with M = 1. However, all integers are representable with M = 2. Thus if M = 1,
+ y)2 + (x - y ) 2 + 5(z + w ) +~ 5(z - w ) =~ 2m. If M = 3, compound with l 2 + 5. l 2 = 6 . Then ( x + 5z 7 + 5 )+' y + 5w + 5 (7)' = 2m (x
(x+)2
(T)2
is an integral representation. For
+ y 2 = 2 + w2 (mod 3),
x2
and we can take
= y = 1, = 0, y = 1,
z
x
and If m
.Y
= 0 (mod 5), say m
=
z
= w = 1 (mod 3), = 0, w = 1 (mod 3).
5m,, from a representation
+ J J ~+ z2 + w2 = 2m,, ~ (x, - ~ J J , + ) ~5(z2 + w 2 ) = lorn,. (2x1 + J J , ) + xy
b
= l,a =
Clearly M 7(12 + 1') For M
=
-7 =
[MI
=
0, 1,2, 3,
(m, 7 ) = 1.
0 is impossible and we may suppose M > 0, since 32
= -1.
2,
x2
+ y2 - 7(z2 + w2) = 2m.
The case x = y (mod 2), z = w (mod 2) is dealt with as for a only consider x = 1, y = 0, z = 1, w = 0, all to mod 2. Since 3' - 7 . l 2 = 2, we need only note ( T3x )2
For M
=
+ 7z 3,
- 7 (x*)2 x2
+ 22 -
+
3y
(-z-)2
+ 7w
-
(T*)
= &
7 y+3cr
+ y 2 - 7(z2 + w2) = 3m.
3. We need
=
m.
172
DIOPHANTINE EQUATIONS
From 22 - 7 . I'
(+2s' + 7z
=
-3, we deduce
-
7
(V)' +
2y
(7)2
(+ + 2w
+ 7w
- 7 y
The brackets are integers since it suffices from x2 consider only x
-m.
+ y2 = z 2 + w2 (mod 3), to
= w E 0 (mod 3), = y = z = w = 1 (mod 3).
x = z = l ,
and
=
y
When m = 0 (mod 7), it suffices to put x = 7 X , y = 7 Y . We take finally a = k 2, b = 5 3, c = 1. It is well known that the result for a = 2, b = 3, c = 1 follows from a = b = c = 1, since x2
+ 2y2 + 3z2 + 6w2 = x' + ( y + z + w)' + ( - y + z + ')M + (z - 242.
We have now the solvable equations with (m, 6 ) = I , I . x2 - 3y2 - 22'
11. x'
+ 3y2
-
+ 6w2 = M m ,
2z' - 6w2 = M m ,
111. x2 - 3y2 + 2 2 - 6w2 = M m . Here 1 M I = 0, 1, 2, 3. Since each of the forms represents - 1, we need only consider M = 0 , 2 , 3. Consider I. Clearly M # 0 since x2 - 2z2 = 0 (mod 3), gives x = 3X, z = 32, and then y = 3Y, w = 3 W . I f M = 3, x = 3 X , z = 3 2 3 X 2 - y 2 - 6 Z 2 + 2w2
and so
and this is I with M = - 1. Suppose next M = 2. Then x
=y
m,
=
(mod 2). If x
2X, y
=
=
2 Y,
2 X 2 - 6 Y 2 - z2 + 3w2 = m, and this is I with M position with Then
= - 1.
If, however, x
=y =
1' - 3.1' - 2.12 + 6.1'
+ 3y + 22 - 6w, = x + z + 3y - 3w,
x2 = x
y2
z2
W '
=
2.
+y =x +y =x
1 (mod 2), apply com-
+ 2w, - z + w, -
22
Hence if w = z (mod 2), x2 = y 2 = z2 = w 2 = 0 (mod 2), and this gives I with M = 1.
19. We can reject
-2
BINARY AND QUATERNARY QUADRATIC FORMS
IV
+ 1 (mod 2) since from I + 6w2 = 2m (mod 4), i.e. m = 0 (mod 2)
=z
2(w - 1)2
-
173
which has been excluded. When m = 0 (mod 6) the representation holds for mi6 and for 6 a n d so for 1H.
When m m. When m
= 0 (mod 2), the representation
holds for j m and for 2 a n d so for
= 0 (mod 3), the representation holds for +m and for 3 a n d so for m. We show now that each of 1 and I1 implies the other. Multiply I with M = 1 by l 2 - 2 . l 2 = - 1. Then
+ 2 ~ - )2(x~ + z ) +~ 3y2 - 6w2 = -m. Since x + 2z = X , x + z = 2 gives a 1-1 correspondence between integer (s
sets (x, z ) and (A’,Z ) , and we have I1 with m replaced by -m. Take finally 111 and multiply by l 2 - 3 . l 2 = -2. Then
~ -)3(x~ - Y ) ~ 4z2
(x - 3
P
where
R
Then x, y , z ,
P
If m If m
E
=
-2m,
3y -t 2z,
Q
= x -
+ 2w,
S
=
P Q - 3RS
Write this a s
IV
=
= x =
-
+ 12w2 = -2m.
x -y
3y - 2z,
x -y -2 ~ ,
will be integers if Q (mod 4),
R = S (mod 4),
P
= R (mod 2).
1 (mod 2), we can solve 111 with P = Q = R = S = 1 (mod 4). M = 1 for +m, for 2 and so for m.
= 0 (mod 2), 111 holds with
REFERENCES 1. B. A. Venkoff. “Elementary Theory of Numbers”. (Russian.) (1937), 51. 2. L. J. Mordell. Solvability of the equation ax2 by2 = p . J. Lond. Math. Soc., 41 (1966), 517-522. 3. L. J. Mordell. The representation of numbers by some quaternary quadratic forms. Actu Arith., 12 (1966), 47-54. 4. J. W. S. Cassels. “An Introduction to the Geometry of Numbers”. SpringerVerlag, Berlin (1959), 31. 5 . L. J. Mordell. Observation on the minimum of a positive quadratic form in 8 variables. J. Lond. Math. Soc., 19 (1944). 3-6.
+
Representation of Numbers by Special Binary Quadratic and Quaternary Quadratic Forms 1. We deal here with the representation of numbers by special quadratic forms in 2 and 4 variables. Such questions are included in the general theory of the representation of numbers by quadratic forms which is discussed in Chapters 18 and 20. Many results however can be found very simply without using the general theory, and this had been long known when a = b = 1. The method now expounded is due to Thue, who applied it in a few instances, as did also Nagell, but it is capable of wider application.
The equation ax2
+ by2 = m.
We suppose that m > 0 and that (ab, m) = 1. Write f ( x , y ) = ax2 by2 = m,
+
We shall suppose that ( x , y ) = 1 and so ( x y , m) = 1. Solvability of equation (l), if ( x , y ) = 1, implies the existence of an integer k such that ak2 b = 0 (mod m). (2)
+
This imposes conditions on a and b. Thus if m is odd, the Legendre symbol ( - a b / p ) = 1 for every prime p dividing m. We first prove the l s 2 Theorem 1 If the congruence
+b
O(modm) is soloable, integers x , y , not both zero, exist such that ak2
ax2
3
+ by2 = M m
(3)
for some integer M with M < 2dab if a > 0, b > 0, and IMI < d/labl if ab < 0. Put x = k X + m Y,y = X . Then by Minkowski's theorem on linear forms in Chapter 5, there exist integers (A', Y ) # (0,O)such that 1x1 <
fdm,
where 1 is any positive constant.
lyl
< I-wm,
19.
165
BINARY A N D QUATERNARY QUADRATIC FORMS
Hence if a > 0, b > 0. f(x,y ) <
if we take l 2
=
(UP+ b/I2)m < 2 d a b m ,
dbz.
If ah < 0,
1f ( x , y)l if we take l 2
=
Further
f ( ~y ),
< max( la1I2rn,lb1Pm) <
d l q m,
dml.
+ m Y)2+ bX2 = (ak2 + b ) X 2 + 2akmXY + am2 Y 2 = 0 (mod m )
=
a(kX
=
Mm.
The problem now is to determine whether M = 1 is possible. Many of the possibilities for M can be eliminated by congruence considerations. Suppose that m is odd. I. congruences mod 4 or mod 8. Thus all to mod 4, a 3 1, b = -1 excludes M = 2 (mod 4) since then x and y are odd and x2 - y 2 I 0 (mod 4). If a = I , b = 1, 2, then solvability for M = 0 (mod 4) implies solvability for M / 4 since now x = y = 0 (mod 2). 11. congruences mod q, q = 3, 5,7,. . ., a prime. If (-ab/q) = - 1, then M $ 0 (mod q). 111. sometimes it may be shown that if M = M,, say, is possible, then M = M 2 is also possible and conversely. For this we use the results,
if then and if then
+ by2 = m, ax: + by? = ml, (axx, + byy# + ab(xyy,- x , ~ =) mm,; ~ x 2 + aby2 = m, as: + by: = m,, a(xx, + byy,)' + b(sy, - ays,)2 = mm,. ax2
(4)
+
(5)
1V. If (m, ab) = 1, and a is square free, then solvability of ax2 by2 = am implies that of x2 aby2 = m and conversely. We suppose for simplicity that m is an odd prime p since solvability when 171 is composite follows easily from the identities in 111. Let us now consider the detailed results. We begin with the case a > 0, h > 0. We take a = 1 and then
+
.'i2
+ by2 = My,
M < 2db,
and (- b/p) = I is the necessary condition for solvability. We write this in a slightly different form, for example, if b e 1 (mod 4), then (- l / p ) ( p / b )= 1,
166
DIOPHANTINE EQUATIONS
and ifb E 3 (mod 4), then (p/b) = I . l f t b = r = 3 (mod 4), then ( 2 / p ) ( p / r )= 1, but if 46 = r = 1 (mod 4), then ( - 2 / p ) ( p / r ) = 1. 2. It will suffice if we give the results for square free b < 20 and give details for a few values of b. M = I i f b = I , 2,3, 7 and (-b/p) = I Take for example b = 7 and so M < 2/28 = 1 , 2 , 3,4, 5 . We can exclude M = 2 , 4 since then x E y = 1 (mod 2 ) and x2 + 7y2 = 0 (mod 8). We can exclude M = 3, 5 since (-7/3) = - 1 = ( - 7 / 5 ) . M = I i f b = 5providedthat(-l/p) = ( p / 5 ) = l . I f ( - l / p ) = ( p / 5 ) = -1, then M = 2 or 3 and either value implies the other. The condition (- 5 / p ) = 1 gives (-lip) = ( p / 5 ) . Here M < d% = I , 2, 3,4. If M = 4, then x = y = 0 (mod 2) and implies M = 1. For this, ( p / 5 ) = 1 and so (- l/p) = 1. Then M = 2 or 3 can arise only when (2p/5) = 1 or (3p/5) = I , i.e. ( p / 5 ) = - 1. Suppose now x2 5y2 = 2p. Multiply by l 2 5. l 2 = 6. Then from equation (4),
+
+
x2
+ 5Y2 = 3p,
+ 5y = 2 x , 3 s = x + 5Y,
where
.x
and so
x 31'
- y = 2Y, =
x-
Y
If x and y are given, x = y (mod 2 ) and so A', Yare integers. If X and Yare given, we can choose Y so that X = Y (mod 3), and then x, y are integers. M = I ifb = 6 provided that ( 2 / p ) = (p/3) = 1 but if ( 2 / p ) = (p/3) = - 1, then 2 . 2 + 3y2 = p . M = 1 ifb = I0 provided that ( - 2 / p ) = ( p / 5 ) = 1 but if ( - 2 / p ) = ( p / 5 ) = - 1, then 2x2 5y2 = p . M = I or 4 jf h = I 1 or 19, but we cannot remove the ambiguity. The solvability of s2 by2 = Mp means that
+
+
s 2
+ sy +
31'2
=p,
A-2
+ xy + 5y2 = p
are both solvable. M = l i f b = 13providedthat(-l/p)=(p/13)= l b u t i f ( - l / p ) = ( p / 1 3 ) = - 1, then A4 = 2 or 7 and M = 2M = 7. M = I i f b = 15 provided that (p/3) = ( p / 5 ) = 1 but if ( p / 3 ) = ( p / 5 ) = - I , then 5 2 + 3y2 = p . Different kinds of results arise when b = 14, 17. Suppose first that b = 14. Then if (2/p) = (p/7) = 1, either M = 1 or 2 2 + 5y2 = p and we cannot remove the ambiguity. If (2/p) = (p/7) = - I , then either M = 3 o r 2x2 5y2 = 3p, and we cannot remove the ambiguity. When h = 17, M = 1 o r 2 if ( - lip) = (p/17) = 1 and we cannot remove the ambiguity. I f (- I/p) = (p/17) = - 1 , M = 3, 6, 7, each implying the other two.
+
19.
I67
B I N A R Y A N D Q U A T E R N A R Y Q ~ J A D K A T I CFORMS
We now consider b = 1 1 in detail. Here A.i < 2d71 = I , 2 , 3, 4, 5, 6. We can exclude M = 2, 6 since s2 + 1 ly2 5 0 (mod 4). If M = 4, the case when x, y are even reduces to M = 1. We show that the case when s and y are both odd, and the cases M = 3, 5, each imply the other two. Suppose s2 + 1 l y 2 = 4p. From l 2 + 1 I . l 2 = 12, we deduce A2
+ l l B 2 = 3p,
where
4A=.~+Ily,4B=.~-y;
3y=A-B,
3.\-=A+lIB.
By choice of sign of y , A and B are integers. If A , B are given, we can take A = B (mod 3), and then x, y are integers. Suppose next that x2 1 1y2 = 5p. From 22 11. l2 = 15, we deduce
+
C2
+ 110'
+
=
3p,
where 5C
=
+
2 . ~ 1 IJ,
5D
=
s - 2 ~ ; 3.u
=
2C
+ 11D,
3 , ~= C - 2 0 .
We can take x = 2y (mod 5) and later C = - D (mod 3), and then have integer correspondences. We show that if these three cases occur, then M = 1 cannot occur and conversely. This is easy t o see from X 2 XY 3Y2 = p, but the result is a particular case of the
+
Theorem 2 The equations ax2
+ by2 = Ip,
ax:
+ by:
=
+
mp, (.u, y)
cannot both be solrable if a > 0, b > 0, (ah, p ) We may take to mod p, x
E
s1 = k y ,
ky,
Since
(asx,
on putting
axx,
=
=
1, (x,, y , ) = 1.
I, I #
ak2 + b
where
in, ltiz
< ah.
= 0.
+ byy,)2 + ah(.uy, = Imp2, + byy, = p X , xy1 - s l y = p Y ,
where X , Yare integers, then X2
+ ahY2 = In?.
Hence Y = 0, i.e. s / y = and then s = ? sl, y = f y , , and this is not so. 3. We turn now to the case ab < 0. We write equation ( I ) in the form xl/jll,
5 2
-
cy2 = Ep,
E
=
f 1.
168
DIOPHANTINE EQUATIONS
Now Some results may be mentioned. M = I if c = 2, 5, 13, 17. We also have here M = - 1 since x2 - cy2 is solvable for these c on noting y = 5 for c = 13. M = k I w h e r e c = 3 , 6 , 7 , 11,14,19,andif
and if
E
c
= 0 (mod 2),
=
=
M
k 1 or
=
* 2 when
=
(5)
=
=
+I,
=
+I.
-1
(6)
=
sign corresponds to M The M = k I i f c = 15and
(t)
(5)
($) k 1 , ($) *1,
1 (mod 2),
c
=
k 1. 1, p
=
f 1 (mod 4).
c = 10 according as
(f)= (5) = I or - I .
The details are comparatively simple except for c Here
19 and so we give them.
=
x2 - 1 9 ~ '= Mp.
I
=
1. We show that any of
4' - 19.1'
=
-3,
We can exclude IM1 = 4 since this reduces to IM M = 1, - 2, - 3 implies each of the other two. For from
X'
- 19y3 = p ,
A' - 19B2 = -3p,
where A
=
4~ - 19y, B
=
x - 4 ~ ; 3~ = -4A
Since we can take A = B(mod 3), M = 1Next M = - 2 tf M = -3. For from S'
-
1 9 ~ '= -2p,
M
+ 19B, =
3y
=
-A
+ 4B.
-3.
5' - 19.1'
=
6,
C' - 1 9 0 ' ~-3p, where 2C
= 5~
- 19y, 2 0
=
x - 5y;
3~
=
5C - 19D, 3y
=
C - 5D.
19.
BINARY AND QUATERNARY QUADRATIC FORMS
169
The result is now obvious since we can take x = y (mod 2), C = - D (mod 3 ) . Similarly, we can show that each of M = - 1 , 2 , 3 implies each of the others. Hence when (- I/p) = I , we can take M = 1 since M = - 1 is excluded by the congruence x2 - 19y2 = - p = - 1 (mod 4). So when ( - I/p) = - I , we have M = - 1. 4. We now consider the representation of numbers m, which we may suppose
square free, by the special quaternary forms
j
= f(x,
y , z , w) = x2
+ bcy2 + caz2 + abw2.
(6)
These forms admit a composition process. Thus if
+ bcy: + caz: + abw:,
fl = f(s1, y,, z,, w,) = xf then where
ffl = f(-hY 2 , z2, )+>2),
fz
=
XZ
= . Y X ~-
(bcyy,
+ CaZ.7.1 + abrtlw,),
(7)
+ xy, + U(ZM’, - WZ1), = + + b(rt’y, yw!,), = wx1 + + c(yzl zy,).
yz = y s , ~2 11’2
XZ,
ZS,
-
XM’,
-
Hence iff represents both m and m,, thenSrepresents mm,. Theorem 3 Integers (s,y , z ,
it,)
# (0, 0, 0, 0 ) exist such that
j ( x , y , z , w) = Mm,
prolitled that the congruence cA2
I M 1 < d2FFC1,
+ bB2 + a E O(modrt7)
(8) (9)
is solaable.for A and B.
It is shown in Chapter 6 that the congruence is solvable if (m, abc) = 1, = c A X + b B Y + m Z , y = B X - A Y + mW,z = X, = Y. From equations (6) and (9),
m f. O(mod4).Puts 11’
,f(s,y,
2 , I+!) =
F ( X , Y, 2,W ) = Mm,
(10)
where M is an integer, and the determinant D of F is a2b2c2m4. There are several methods for estimating a minimum value p o r lower bound for a quaternary form F ( X , Y , Z, W ) of determinant D. The most elementary is to use Hermite’s* lpl < (4)3’2qrq, and so M < 1 . 5 4 d m l . Next perhaps is the application of Minkowski’s theorem on linear forms in Chapter 5. Thus integers (X,Y , Z , W ) # (0, 0, 0, 0) exist such that, if 6 = q/labcl, then f 61x1 < 2- labanl, Sly1 < SIzI < djhnll, 61wl < dlGl. This gives IMI < 4dlad;l. The estimate (8) follows from the well known 4 more precise results of Korkine and ZolotarefP lpl < 2/4\01. These ~~
~~
,I%[’%
170
DIOPHANTINE EQUATIONS
estimates are usually given for positive definite forms, but they also hold for indefinite forms although the results are not very sharp. We commence with the case c = 1 and so
f = x2 + by2 + az2 + abw2 = Mm, IMI
< dm,
+
(1 1)
and A 2 + bB2 a = 0 (mod m), (m, ab) = 1 (12) The classical case is a = b = 1, Hermite’s result gives M < 1-54, i.e. M = 1. Minkowski’s estimate gives A4 < 4 and so M = I , 2, 3. The case M = 2 can be written as
(F)’ + (y)’ + + + (T) ( Tz ) z
U’
Z-MM”
=
m,
since either x, y , z , M: are all even or odd or only two odd, say x, y. The case M = 3 can be written as
(“ + ;
+;
+ 3”
1’
)Z - 2)Z (x - w)’ (x -y = 171, (y M’ since we may suppose s = 0 (mod 3), y = z E w = 1 (mod 3). The method shows that m is representable by f in equation ( 1 I ) when b = I , a = &3, 5 5 , -7, -11 and also when a = 5 2 , b = 3, -3. It will suffice to take a few of these cases.
+; +
b
=
If M If x
l,a =
=
+
M <
3
2,
s2
+
d6 = 1, 2,
(m,3)
+
=
+ y2 + 3(z2 + w2) = 2m.
1.
(13)
= y (mod 2), then z = w (mod 2), and
and this is the case M = 1. If x = y 1 (mod 2), then z
+
1
= 14’
+ 1 (mod 2), and so from equation (13)
+ 3 = 2m (mod 4),
m
= 0 (mod 2).
Hence M = 1 if m = 1 (mod 2). I f m = 0 (mod 2), the argument applies to t m , and since 2 is representable b y f so is m . Finally if m z 0 (mod 3), +m is representable by f and so is 3 and hence also m. b = I , a = -3 (m,3) = 1. ] M I = 0, 1,2, Since - 1 is representable, we need only consider M 3 0. Clearly M = 0 is impossible since this requires x 2 y = 0 (mod 3) and then z = w = O(mod 3). If M = 2, the argument above applies when x = y (mod 2), z = iti (mod 2). For .Y E y 1 (mod 2), z = it’ 1 (mod 2), we need only consider
+
x
= 1 (mod 2),
y
= 0 (mod 2),
z
+
= I (mod 2),
w
= 0 (mod 2),
19.
171
BINARY AND QUATERNARY QUADRATIC FORMS
Since
12
-
3.12 =
-2,
we have
(T)
( x + 3z 7 - 3 ) +' ( y + 3w 7 -3 y) + w 2 =' -m, (x+)2
and this is the case M = - 1. The case m = 0 (mod 3 ) is dealt with as above. b = I , a = 5, then (m, 5) = 1 and this case is of some interest. Here
x2
+ y 2 + 5(z2 + w2) = Mm,
M
1,2, 3.
=
Clearly M # 1 for all m since 3 is not representable. It can be shown from the theory of the ternary quadratic that all integers > 3 are representable with M = 1. However, all integers are representable with M = 2. Thus if M = 1,
+ y)2 + (x - y ) 2 + 5(z + w ) +~ 5(z - w ) =~ 2m. If M = 3, compound with l 2 + 5. l 2 = 6 . Then ( x + 5z 7 + 5 )+' y + 5w + 5 (7)' = 2m (x
(x+)2
(T)2
is an integral representation. For
+ y 2 = 2 + w2 (mod 3),
x2
and we can take
= y = 1, = 0, y = 1,
z
x
and If m
.Y
= 0 (mod 5), say m
=
z
= w = 1 (mod 3), = 0, w = 1 (mod 3).
5m,, from a representation
+ J J ~+ z2 + w2 = 2m,, ~ (x, - ~ J J , + ) ~5(z2 + w 2 ) = lorn,. (2x1 + J J , ) + xy
b
= l,a =
Clearly M 7(12 + 1') For M
=
-7 =
[MI
=
0, 1,2, 3,
(m, 7 ) = 1.
0 is impossible and we may suppose M > 0, since 32
= -1.
2,
x2
+ y2 - 7(z2 + w2) = 2m.
The case x = y (mod 2), z = w (mod 2) is dealt with as for a only consider x = 1, y = 0, z = 1, w = 0, all to mod 2. Since 3' - 7 . l 2 = 2, we need only note ( T3x )2
For M
=
+ 7z 3,
- 7 (x*)2 x2
+ 22 -
+
3y
(-z-)2
+ 7w
-
(T*)
= &
7 y+3cr
+ y 2 - 7(z2 + w2) = 3m.
3. We need
=
m.
172
DIOPHANTINE EQUATIONS
From 22 - 7 . I'
(+2s' + 7z
=
-3, we deduce
-
7
(V)' +
2y
(7)2
(+ + 2w
+ 7w
- 7 y
The brackets are integers since it suffices from x2 consider only x
-m.
+ y2 = z 2 + w2 (mod 3), to
= w E 0 (mod 3), = y = z = w = 1 (mod 3).
x = z = l ,
and
=
y
When m = 0 (mod 7), it suffices to put x = 7 X , y = 7 Y . We take finally a = k 2, b = 5 3, c = 1. It is well known that the result for a = 2, b = 3, c = 1 follows from a = b = c = 1, since x2
+ 2y2 + 3z2 + 6w2 = x' + ( y + z + w)' + ( - y + z + ')M + (z - 242.
We have now the solvable equations with (m, 6 ) = I , I . x2 - 3y2 - 22'
11. x'
+ 3y2
-
+ 6w2 = M m ,
2z' - 6w2 = M m ,
111. x2 - 3y2 + 2 2 - 6w2 = M m . Here 1 M I = 0, 1, 2, 3. Since each of the forms represents - 1, we need only consider M = 0 , 2 , 3. Consider I. Clearly M # 0 since x2 - 2z2 = 0 (mod 3), gives x = 3X, z = 32, and then y = 3Y, w = 3 W . I f M = 3, x = 3 X , z = 3 2 3 X 2 - y 2 - 6 Z 2 + 2w2
and so
and this is I with M = - 1. Suppose next M = 2. Then x
=y
m,
=
(mod 2). If x
2X, y
=
=
2 Y,
2 X 2 - 6 Y 2 - z2 + 3w2 = m, and this is I with M position with Then
= - 1.
If, however, x
=y =
1' - 3.1' - 2.12 + 6.1'
+ 3y + 22 - 6w, = x + z + 3y - 3w,
x2 = x
y2
z2
W '
=
2.
+y =x +y =x
1 (mod 2), apply com-
+ 2w, - z + w, -
22
Hence if w = z (mod 2), x2 = y 2 = z2 = w 2 = 0 (mod 2), and this gives I with M = 1.
19. We can reject
-2
BINARY AND QUATERNARY QUADRATIC FORMS
IV
+ 1 (mod 2) since from I + 6w2 = 2m (mod 4), i.e. m = 0 (mod 2)
=z
2(w - 1)2
-
173
which has been excluded. When m = 0 (mod 6) the representation holds for mi6 and for 6 a n d so for 1H.
When m m. When m
= 0 (mod 2), the representation
holds for j m and for 2 a n d so for
= 0 (mod 3), the representation holds for +m and for 3 a n d so for m. We show now that each of 1 and I1 implies the other. Multiply I with M = 1 by l 2 - 2 . l 2 = - 1. Then
+ 2 ~ - )2(x~ + z ) +~ 3y2 - 6w2 = -m. Since x + 2z = X , x + z = 2 gives a 1-1 correspondence between integer (s
sets (x, z ) and (A’,Z ) , and we have I1 with m replaced by -m. Take finally 111 and multiply by l 2 - 3 . l 2 = -2. Then
~ -)3(x~ - Y ) ~ 4z2
(x - 3
P
where
R
Then x, y , z ,
P
If m If m
E
=
-2m,
3y -t 2z,
Q
= x -
+ 2w,
S
=
P Q - 3RS
Write this a s
IV
=
= x =
-
+ 12w2 = -2m.
x -y
3y - 2z,
x -y -2 ~ ,
will be integers if Q (mod 4),
R = S (mod 4),
P
= R (mod 2).
1 (mod 2), we can solve 111 with P = Q = R = S = 1 (mod 4). M = 1 for +m, for 2 and so for m.
= 0 (mod 2), 111 holds with
REFERENCES 1. B. A. Venkoff. “Elementary Theory of Numbers”. (Russian.) (1937), 51. 2. L. J. Mordell. Solvability of the equation ax2 by2 = p . J. Lond. Math. Soc., 41 (1966), 517-522. 3. L. J. Mordell. The representation of numbers by some quaternary quadratic forms. Actu Arith., 12 (1966), 47-54. 4. J. W. S. Cassels. “An Introduction to the Geometry of Numbers”. SpringerVerlag, Berlin (1959), 31. 5 . L. J. Mordell. Observation on the minimum of a positive quadratic form in 8 variables. J. Lond. Math. Soc., 19 (1944). 3-6.
+