A R K I V F O R MATEMATIK Band 3 nr 6
R e a d 11 N o v e m b e r 1953
On the Diophantine equation xZ+ 8D =yn
By TRYGVE ~TAGELL
§ ] .
I n a previous p a p e r I I showed t h a t t h e D i o p h a n t i n e e q u a t i o n
(1) x~ + 8 = y n (n>=3)
has no solution in positive integers x a n d y w h e n n is n o t a prime = + 1 (mod 8).
I f n is a prime -: +_ 1 (mod 8), there is at m o s t one solution in p o s i t i v e integers.
I t is, however, possible t o o b t a i n t h e following i m p r o v e m e n t of this r e s u l t : T h e o r e m i . The Diophantine equation (1), where n is an integer > 3 , has na solution in positive integers x and y.
T h e proof will be given in § 5.
I n this p a p e r we shall e x a m i n e t h e more general e q u a t i o n
(2) x 2 + 8 D = yn,
where D is a square-free, o d d integer > 1, a n d w h e r e n is a n integer ~ 3.
W e begin b y p r o v i n g t h e following l e m m a :
Lemma 1. The equation (2) has no solution in even integers x and y i/ n > 4.
I ] n = 3 and i/ the number o/ ideal classes in the quadratic field K ( V - 2 D ) is not divisible by 3, the equation (2) is solvable in even integers x and y only when D = 6 a 2 T 1, a integer; corresponding to this value o/ D there is the single integral solution y = 16 a * ~ 2.
Proof. L e t x, y be a solution of (2) in integers. I f x is even, y is so. T h e n yn is divisible b y 8. H e n c e b y (2) x is divisible b y 4. Since D is odd, y~ m u s t be divisible b y e x a c t l y 8, a n d this implies n = 3. If we p u t x = 4 x 1 a n d y = 2 Yl, we get
(3) (2 xl) 2 + 2 D = 2 y~.
T h e ideal factors ( 2 x 1 + V = 2 D ) a n d ( 2 x I - ~ / - 2 D ) of t h e left-hand side h a v e t h e greatest c o m m o n divisor (2, V - 2 D ) . H e n c e it follows f r o m (3)
1 See NAOEr.L [1], § 2. Figures in [ ] refer to the Bibliography at the end of this paper.
*r. NAGELL, On the Diophantine equation xS + 8 D =yn
(4) (2 x I + U - 2 D) = (2, ] / ~ 2 D) i a,
w h e r e i is a n ideal w i t h t h e n o r m Yl in K ( ] / ~ 2 D). Since (2, V - 2 D) s = (2),
w e g e t
(5) (2 x 1 + V - 2 D) s = (2) i s.
T h u s i e is a principal ideal. Since, b y hypothesis, t h e class n u m b e r is n o t divisible b y 3, it is e v i d e n t t h a t i S is a principal ideal. T h e n it follows f r o m (5) (6) ( 2 x 1 + V - 2 D ) s = 2 ( u + v l / - 2 D ) a,
w h e r e u a n d v are r a t i o n a l integers, such t h a t
(7) y~ = (N i) ~ = u 2 + 2 D v s.
u is odd, since Yl is so. I t follows f r o m (6) t h a t (8) u + v I / - 2 D = (a 1/2+ b V-~-- D) ~,
w h e r e a a n d b are r a t i o n a l integers. C o m b i n i n g this e q u a t i o n w i t h (6) we g e t (9) x~ ]/2 + ~ / - D = (a V2 + b 1 / - D) a,
w h e n c e
T h i s implies b = ± 1 a n d (10)
T h e n we g e t f r o m (7) a n d (8) (11)
a n d f r o m (9) (11')
l = 6 a S b - D b a.
D = 6 a ~ T 1.
yl = N i = 2aS + DbS = 2a2 + D = 8 a 2 - ~ 1,
x t = 2 a 3 - 3 D a b s= -- 1 6 a a + 3 a .
§2.
W e shall n o w consider e q u a t i o n (2) for a n o d d solution x. L e t n be t h e p o w e r of a n o d d p r i m e q, t h u s n = q ~. F u r t h e r w e suppose t h a t t h e n u m b e r of i d e a l classes in t h e q u a d r a t i c field K ( 1 / - 2 D) is n o t divisible b y n.
W h e n x is odd, y is also odd, a n d t h e ideal f a c t o r s @ + 2 I / - 2 D ) a n d ( x - 2 ~ / - 2 D) of t h e l e f t - h a n d side of (2) are r e l a t i v e l y prime. H e n c e
(12) ( x + 2 V - 2 D ) = i " ,
ARKIV FOR MJkTEMATIK. B d 3 nr 6 where i is a n ideal. I f the class n u m b e r h in K ( ~ / - - - ~ ) is divisible b y q~ (0 _~ fl < ~) a n d not b y
qt~+l, there
exist two rational integers / and g such t h a tT h e n b y (12) we get the following equivalence
~q~ ~ ~ rq~ N 1.
H e n c e we obtain from (12)
x+ 2 ~ = (u+v V~-2D) q,
where u and v are rational integers, such t h a t y~ = (u ~ + 2 D v~) q.
u is odd since x is so. Then, equating the coefficients of V-~-2D, we get the relation
(13) 2 = ~ . U q-2k-1 V 2 k + l ( - - 2 D ) k.
~.=0 2 . 1
F r o m this equation it is obvious t h a t v is a divisor of 2 a n d t h a t
quq-lv
is even. Hence v = ± 2 since q and u are odd. All t h e t e r m s on t h e right-hand side in (13) are divisible b y q, except the last t e r m (for k = ½ (q--1)). Thus we get, if D is n o t divisible b y q,- 2 D
2~vq ( - 2 D)½(q-1)=-v(--~ - )
( m o d q ) , whenceI f D is divisible b y q, equation (13) is impossible.
Then, on dividing (13) by v, we have
(14)
( ~ ) = ½(;~=2) (2 :+ l) Uq-2k-l (-- S D) k.
Taking equation (14) as a congruence modulo 8 we get
( - ~ ) =qu "-~-q
(mod 8), (15)whence it follows
q - - ± 1 (mod 8).
Hence, taking in consideration L e m m a 1, we have t h e following result:
T h e o r e m 2.
Let n be the power el an odd prime q, n ~ 3, and suppose that
the class number in K (V-~-D) is not divisible by n.
T. NAGELL, On the Diophantine e~uatien x2 ~ - 8 D = y n
1] q=-- q-3 (mod 8), the Diophantine equation (2) has no solution in integers x a n d y, apart from the case when n = 3 and x and y are even. Likewise, i/ D is divisible by q, equation (2) has no integral solution.
We m a y also state
L e m m a 2. Let n be the power of a prime q=~ ~ 1 (rood 8), and suppose that the class number in K ( V - 2 D ) is not divisible by n.
I f the Diophantine equation (2) is solvable in integers x and y, we must have
n
yq = un + 8 D, where u is an odd integer satisfying equation (14).
§a.
Now suppose t h a t the prime q in (14) is = +__1 ( m o d 8 ) . If we p u t X = u ~ and Y = - 8 D , the right-hand side of (14) becomes a form of the degree
½ ( q - 1 ) in X and Y with integral coefficients. B y the theorem of EIs~,~ST~I)r it is obvious t h a t this form is irreducible. Hence, according to a famous theorem of Taut., equation (14) holds only for a finite number of integral values X and Y. Thus we have proved:
T h e o r e m 3. Let n be the power o/ an odd prime =-- +_ 1 (rood 8), and suppose that the class number in K ( V - 2 D ) is not divisible by n. .For a given n~_7, there is only a ]inite number of square-free odd integers D ~ I , such that the Diophantine equation (2) is solvable in integers x and y.
When q ~ - 1 (mod 8) it follows from (15) - 2 D
W h e n q - + 1 (rood 8) it follows - 2 D
Hence, in both cases D must be a quadratic residue modulo q. Thus we can state T h e o r e m /~. Let n be the power o/ an odd prime q ~ ± 1 (mod 8), and suppose that the class number in K ( ~ - - 2 D ) is not divisible by n. I f D is a quadratic non-residue modulo n, the Diophantine equation (2) has no solution in integers x and y.
ARKIV FOR MATEMATIK. B d 3 n r 6
§ 5 . Now we suppose t h a t D - 1 (mod 3).
If q - - 1 (rood 8) it follows from (14)
~(q--1) --1----
k~O ( 2 : + 1) uq-~k-1 (mod 3).
This is impossible when u is divisible b y 3, since, in t h a t case, the right-hand side is - 1 (rood3). If u is not divisible by 3, we get
B u t this congruence is impossible since the value of the right-hand side is 2 q--1 and thus : 1 (mod 3).
If q--1 (rood 8) it follows from (14)
½(q--l)
(16) q - - l + q ( u q - ~ - l ) = --
k = l ( 2 : + 1) uq-~k-1 ( - 8 D)~"
Suppose q - l = 2 r q l where ql is odd. Then U q - l - 1 is divisible by 2 r+2. The general term in the right-hand sum in (16) m a y be written
q ( q - 1 ) ( : f f ) u q _ ~ k _ l ( _ 8 D ) ~ .
(17) 2 k (2k-+]) 1
Here the numerator is divisible by 2 r+3k. The denominator is divisible b y a power of 2 which is g 2 k . Since for all k=>l
2 a ~ = 8 ~ > 4 k ,
we conclude t h a t the number (17) is divisible at least by 2 r+l. Hence equa- tion (16) is impossible, for q - 1 is divisible by 2 r but not by 2 "+1.
Thus we can state
T h e o r e m 5. Let n be the power o/ an odd prime q-++_1 (mod 8), and suppose that the class number in K ( V - 2 D ) is not divisible by n. I / D - 1 (mod3), the Diophantine equation (2) has no solution in integers x and y:
This result is contained in the more general
T h e o r e m 6. Let n be an odd integer > 3, and suppose that the class number in K ( | / ~ ) is not divisible by n. I / D = - 1 (mod 3) the Diophantine equation (2) has no solution in integers x and y.
Proo/. Suppose t h a t equation (2) is solvable in integers x and y. There must exist a prime factor q of n with the following property: q~ is a factor of n b u t
T. rCXCELL, On the Diophantine equation x~ + 8 D = y n
not of the class number h. Let us p u t m = q ~, n = m r and z = y r. Then the equation
(2') x ~ + 8 D = z m
should be solvable in integers x and z. B u t by Theorem 2 this is impossible when q ~ +_3 (mod 8) and m = q ~ > 3 . When m = q ~ = 3 , it follows from L e m m a 1 and Theorem 2 t h a t
z = y r = 16a2T 2;
b u t this is impossible since r = ~ > 1. n
When q ~ ± 1 (rood 8) equation (2') is impossible in virtue of Theorem 5.
I n the special case D = 1 we easily get Theorem 1. I n fact, the class n u m b e r in K (V----2) is = 1. The equation
x ~ + 8 = y~
is possible only for I 1= 1, I v l = a By r mma 1 the equation
x ~ + 8 = ya is satisfied only for x = 0, y = 2.
§ 6 . We shall prove the following theorem:
T h e o r e m 7. Let n be the power o I an odd prime q = ± 1 (mod 8), and suppose that the class number in K (V~-2D) is not divisible by n. T h e n the Diophantine equation (2) has at most one solution i n positive integers x and y.
Prool. Suppose t h a t equation (14) was satisfied for two values u and u 1 ( u # ± % ) . Thus
( ~ ' - ~ ) ~½(;~:)(2:~_l)q~bq-2k-l(--UD) k~
Subtracting this equation from equation (14) we get, on dividing b y u 2 - u ~ :
uq-1 -- uq-1 ½(q--3) (
) u q - 2 k - l - - u q-2k-1
(18) q u 2 - u ~ k=~l 2 : + 1 u ' - u ~ (--8D)k"
We need the following lemma:
L e m m a 3. Suppose that m = 2 ~" r, where m, l ~ and r are positive integers, r odd.
~uppose /urther that u and u 1 are odd integers u # +_%. T h e n the integer
U rn -- U~
U 2 -- ~21
is divisible by exactly 2 ~-1 and not by 2 ~'.
ARKIV FOR MATEMATIK. Bd 3 nr 6 Proo[. The l e m m a is true for m = 2 , independently of the value of r. For,
~ince r is odd, t h e n u m b e r
u~r--u~ _U~,_2+U~_4U~+... +U~-~
U 2 - - U ~
is odd. Suppose t h a t the l e m m a is true for the even exponent m. T h e n we shall show t h a t it is also true for the exponent 2 m. I n fact, we have
u~m-u~ ~ u~_u~
u~_u~ (u~+u?) -
U 2 -- U~ ~
a n d urn+ u~, being the s u m of two odd squares, is even b u t n o t divisible b y 4.
Thus L e m m a 3 is established b y induction.
I t is easy to see t h a t equation (18) is impossible when q-= - 1 (rood 8). For, b y L e m m a 3, the left-hand side of (18) is odd in this case. B u t the right- h a n d side is divisible b y 8.
Suppose n e x t q - 1 (mod 8) a n d q - 1 = 2 s t , where r is odd a n d # ~_ 3. Then, b y L e m m a 2, the left-hand side of (18) is divisible b y 2 ~-1 and not b y 2 ~'.
T h e general t e r m in (18) m a y be written
( : / 2 )
q(q-1)
23k Uq-2k-1--U~ -2k-11 2 k ( 2 k + 1) u~-u~ (-D)~"
Since for all k ~ l
23k > 2 It,
¢his n u m b e r is divisible at most b y 2 ~. Hence the right-hand side of (18) is divisible b y 2 ~. B u t we h a v e just shown t h a t the left-hand side of (18) is divis- ible b y 2 "-1 a n d not b y 2 ~. Thus equation (14) is satisfied b y at m o s t one v a l u e of u S. The corresponding value of y is given b y the relation
(19) yn = (u 2 + 8 D) q.
T h i s proves T h e o r e m 7.
§ 7 . F u r t h e r we prowe
T h e o r e m 8. Let n be an odd integer > 3 , and suppose that n and the class n u m b e r in K ( ~ 2 D ) are relatively prime. I [ the Diophantine equation (2) has a solution in integers x and y, n is a prime - ± 1 (mod 8).
Proo[. Suppose t h a t n is divisible b y a prime q = --+3 (rood 8). P u t
n
Z = yq a n d consider the equation
x2 + 8 D = z q.
T. NAGELL, On the Diophantine equation x2 + 8 D =yn
The class n u m b e r h in K ( V - 2 D ) is not divisible b y q. I f q = 3 , we get b y L e m m a 1
z = 1 6 a 2 ~ 2 = y ~.
B u t this is clearly impossible since ~ > 1. I f q > 3, it follows from T h e o r e m 2 t h a t equation (2) is impossible.
Hence n is a p r o d u c t of primes - ± 1 (mod 8). L e t q be the least of these primes a n d suppose t h a t n > q. P u t
n_
z = yq a n d consider the equation
x2 T $ D = z q.
Since t h e class n u m b e r h is not divisible b y q, it follows b y L e m m a 2 t h a t
z = u 2 + 8 D ,
where u is an odd integer satisfying equation (14). _n is divisible b y a prime p q
which is _~q a n d ~- ± l ( m o d 8 ) . Now p u t
n Z 1 = yV q and consider the equation
u ~ + 8 D = z ~ .
Since the class n u m b e r h is not divisible b y p, it follows b y L e m m a 2 t h a t z1= u~ + 8 D,
where u 1 is an odd integer. Hence we have
n
(20) z = yq = (u~ + 8 D) v >= (1 + 8 D) q.
l~rom equation (14) it follows t h a t
(8 D ) ½ ' q - 1 ) - ( ~ ) ( m o d u~q),
w h e n c e
u2q ~ (8 D) ½(q-l) + 1.
Thus we get
z = u S + 8 D _-< _1 [(8 D) ½(q-l) -t- l] -~ 8 D.
q
B u t this contradicts the inequality (20). I n fact, it is easily seen t h a t for all D>= 1 and all q ~ 7 , we h a v e
ARKIV FOR MATEMATIK. B d 3 nr 6 (1 + 8 D ) ~ > q [(8 D) ½(q-l) + 1] + 8 D. 1
H e n c e n m u s t be a prime = _+ 1 (mod 8), a n d T h e o r e m 8 is proved.
§8.
F i n a l l y we p r o v e
T h e o r e m 9. Let n be an odd integer > 3, and let D be a positive integer of the form
(21) D = ~ (y= - x2),
where x and y are odd integers. Then there exists a number D o such that the class number in the imaginary quadratic /ield K ( ~ 2 D) is divisible by n for all square- ]ree D >= Do.
Proof. Suppose t h a t t h e class n u m b e r is n o t divisible b y n. T h e n there exists a prime f a c t o r q of n w i t h t h e following p r o p e r t y : q~ is a f a c t o r of n b u t n o t of t h e class n u m b e r . L e t us p u t m = q ~, n = m r a n d z = y r. T h e n it follows f r o m (21)
(22) x 2 + 8 D = z m.
B u t b y T h e o r e m 2 this relation is n o t possible for integral values of x a n d z w h e n q - _+ 3 (rood 8) a n d m = q ~ > 3. W h e n m = q~= 3, it follows f r o m T h e o r e m 2 t h a t (22) is n o t possible for even x a n d z. W h e n q ~ + 1 (rood 8), in v i r t u e of T h e o r e m 3, t h e relation (22) is possible only for a finite n u m b e r of values D.
This proves T h e o r e m 9.
Remark. I t m a y be shown t h a t there are infinitely m a n y positive a n d square-
~ree integers D of t h e f o r m (21); c o m p a r e [2], § 2.
§ 9 .
There are several similar results on other D i o p h a n t i n e e q u a t i o n s of t h e t y p e
(23) x 2 + B = yn,
where B a n d n are positive integers, n odd a n d t h a t t h e e q u a t i o n
x 2 + l = y ~
> 3. T h u s LnBESOVE s h o w e d
h a s no solution in integers x a n d y for x # 0; see [3].
I n a previous p a p e r I e x a m i n e d e q u a t i o n (23) w h e n B is a positive square- free integer which is either = 1 or = 2 (mod 4), a n d s h o w e d h o w all integral solutions m a y be f o u n d in m a n y cases; see [4], § 2. E x a m p l e : F o r B = 5 a n d n > 3 e q u a t i o n (23) has no integral solution.
T. NAGELL, On the Diophantine equation x 2 + 8 D = y n
LJU~GaREN has treated the case in which B is a positive square-free integer of the form
B = 1 + 22m+l (2 h - 1),
where m and h are positive integers; when the class number in the field K ( V - B ) is not divisible b y n, he showed t h a t equation (23) has no integral solution; see [5] and [6]. Example: For B = 9 and n > 3 equation (23) cannot, be satisfied b y a n y integers x and y.
E q u a t i o n (23) is a special case of the Diophantine equation
(24) ax~ + bx + c = d y ",
where the left-hand side is an irreducible polynomial of the second degree, having integral coefficients; d is an integer # 0. I t was shown by THU• that, this equation has only a finite number of integraI solutions x, y, when n > 3;.
see [7]. This result was subsequently discovered again b y LA~DA~ and 0STROWSKI;.
see [8]. However, no general method is known for determining all integral solu- tions x and y of a given equation of the form (24).
B I B L I O G R A P H Y
[I]. T. NAOELL, Verallgemeinerung eines F~RMAT-schen Satzes, A r c h i v der M a t h e m a t i k , B d V~
S. 53, Ziirich 1954.
[2]. Z u r A r i t h m e t i k der Polynome, B e r i c h t e d. m a t h e m . Seminars d. Hamburgischen- U n i v e r s i t ~ t , B d I, § 2, S. 185, H a m b u r g 1922.
[3]. V. A. LEBESGUE, Sur l'impossibilit6 e n n o m b r e s entiers de l ' ~ q u a t i o n x m = y ~ + 1, N o u - velles Annales de M a t h 6 m a t i q u e s (1), t o m e 9, p. 178, P a r i s 1850.
[4]. T. I~AO~LL, Sur l'impossibilit~ de quelques 6quations ~ deux ind6termin6es, Norsk M a t e - m a t i s k F e t c h i n g s Skrifter, Ser. I, Nr. 13, K r i s t i a n i a 1923.
[5]. W. LJUNGGRr:~, On t h e D i o p h a n t i n e e q u a t i o n x 2 + p ~ = y n, Kong. Norske V i d e n s k a b e r s Selskab, F o r h a n d l . B d X V I , Nr. 8, T r o n d h j e m 1943.
[6]. - - O n t h e D i o p h a n t i n e e q u a t i o n x ~ + D = y n , Kong. Norske V i d e n s k a b e r s Selskab, For- h a n d l . B d X V I , Nr. 23, T r o n d h j e m I944.
[7J. A. THUE, l~ber die U n l S s b a r k e i t der Glcichung a x ~ + b x + c = d y n i n grossen g a n z e n Z a h l e n x u n d y, A r c h i v for M a t h e m a t i k og N a t u r v i d e n s k a b , B d K X X I V , K r i s t i a n i a 1916~
[8]. E. LANDAU a n d A. OsT~owsxI, On t h e D i o p h a n t i n e e q u a t i o n a y ~ + b y + c = d x n, P r o - ceedings of t h e L o n d o n M a t h e m a t i c a l Society, Set. 2, Vol. 19, L o n d o n 1919.
Tryckt den 13 april 1954
Uppsala 1954. Almqvist & Wiksells Boktryckeri AB
