PROCEEDINGS OF THE
„„„
AMERICAN MATHEMATICAL SOCIETY
Volume 69, Number 2, May 1978
THE SOLUTION OF 3y2 ± 2" = x3
STANLEY RABINOWITZ1
Abstract.
The diophantine equation
(»)
3y2 + 2"y = x3,
with y = ± 1
is solved.
Let 9 = 21/3, where 0 E Reals. Then fi = {a + b0 + c92\a, b, c E Z) is
the ring of integers of Q(9), -1 + 6 is the fundamental unit of Ü and ñ is a
unique factorization domain (U.F.D.) (see e.g. [3]).
All English letters (except Z, Q and N) will represent elements of Z and all
lower case Greek letters elements of ñ.
a\aß and (a, ß)Q are read (respectively) as "a divides ß in ti" and "the
greatest common divisor of a and ß in ß".
Table I is taken from [1].
Table I
The solution of y2 + v = x3 for some values of v.
v: Solutions (x, |y|>
3: no solutions
- v: Solutions (x, |y|>
3: <1, 2>
24: no solutions
27:<3,0>
54: <7, 17>
108: no solutions
216: <6,0>, <10,28>, <33, 189>
24: < - 2, 4>, <1,5>, <10,32>,<8158,736844)
27:<-3,0>
54: <3,9>
108: < - 3, 9>, < - 2, 10), <6, 18>,<366,7002>
216: < - 6, 0>
432 : <12, 36>
432 : no solutions
Note that if (*) holds, then n > 0, since 2" = y(x3 — 3y2).
Proposition
1. // (*) holds with n = 3k and xy odd, then <y, 77,x, |y|) =
<1,3, 11,21>.
Proof. Since n = 3k, (9y)2 + 27 • 23*y= (3x)3. Using Table I we may
assume that k > 1. By (*), 3y2 = ab, where a = x - 2ky and b = x2 + 2*yx
+ 22k. Since x is odd, b is odd. b > 0 [3, Lemma 1] and therefore a > 0.
Received by the editors October 18, 1976.
AMS (MOS) subject classifications(1970). Primary 10B10.
Key words and phrases. Ring of integers, norm.
'This material forms a part of the author's doctoral dissertation ("On Mordell's equation
y2 + k = x3, with 7c= ±2"3""') completed in 1971 at the City University of New
York/Graduate
Center.
© American Mathematical
213
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Society 1978
214
STANLEYRABINOWTTZ
a = x -2ky
= x3 - 23A:y= 3y2 = 0 (Mod 3).
Hence x = 2*y (Mod 3), which implies that
b = x2 + 2*yx + 22* = 22* + 22k + 22k = 0 (Mod 3).
Thus 9|a¿? = 3y2, implying that 3\y. Also (a, b)\3 • 22k [3, Lemma 2]. Since b
is odd, (a, b) = 3. Since 3(y/3)2 = (a/3)(Z?/3), [3, Lemma 3] proves that
<a/3, ¿?/3> = (r2, 3s2)
or
<3r2, 52>.
If <a/3, Z?/3> = <r2, 3s2>, then x = 3r2 + 2*y and we obtain
3(i2 - 2V2 - Ia) = 22k,
which cannot hold. Therefore <a/3, Z?/3> = <3r2, s2>. Hence s is odd, x =
9r2 + 2*y, and
27r4 + 9 • 2*yr2 + (22* - s2) = 0.
108j2 - 27 • 22* = d2 [3, Lemma 4]. Consequently 3322|a"2,since k > 1. Thus
d = 18Z>and s2 - 3D2 = 22k~2. Therefore D is odd and it follows that
52 - 3D2 = -2(Mod 8). But 4|22*"2.
Proposition 2.1f(*) holds with n = 3k + 1 and xy odd, then
<y,«,x, |y|> = <-l, 1, 1, 1>,<-1,7, -5, 1> or <-1, 13, 1915,48383).
Proof. If k = 0, then (9y)2 + 54y = (3x)3. We may assume that k > 0 by
using Table I.
(x, 3) = 1, for otherwise 3|2". Thus by [3, Lemma 5],
3y2 = x3 _ 23*+iy = ± i ± 2 (Mod 9).
Hence (y, 3) = 1. By (*), 3y2 = aß, where a = x - 2kyBand ß = x2 + 2ky9x
+ (2k9)2.ß > 0 [3, Lemma 1] and thus a > 0. (a, ß)a\a3(2k9)2 [3, Lemma 2].
Since (aß, 2)ß = 1 [3, Lemma 6], (a, ß)a\a3 = (1 + 9)3(- 1 + 9).
By [4, (13), p. 132]and [2, Theorem 2-11, p. 48] the norm N of fi (over Z) is
A(a + b9 + c92) = a3 + 2b3 + 4c3 - 6abc.
Now A(l + 9) = 3. Thus 1 + 9 is a prime of ß [2, Theorem 2-12, 2-15, pp.
49, 53].Also
x + 2*y = x3 + 23*y = x3 - 23*+1y = 3y2 = 0
(Mod 3).
Let w = (x + 2*y)/3. Then a = (1 + 0)(w - 2ky - w9 + w92). Therefore
1 + 9\aa. If (1 + f?)2|ßa, then 9 = A((l + 0)2)|A(a) = 3y2, contradicting
( v, 3) = 1. Since Q is a U.F.D. and -1 + 9 is a unit of 0, (a/(\ + 9), ß)u =
1. Also
((I + 9)yf= (a/ (\ + 9))(ß/ (-\ + 9)).
a/(\ + 9) = p(a + b9 + c92)2,where p = 1 or -1 + 9 [3, Lemma 3]. We
may assume that a - b > 0, since (a + b9 + c92)2 = (- a - b9 - c92)2.
If jn = 1, then since 1, 9 and 92 are linearly independent over Q,
x = a2 + 4bc + 2b2 + 4ac
and
-2*y = a2 + 4bc + 2c2 + 2aZ?.
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THE SOLUTIONOF 3y2 ± 2" = X3
215
The first equation implies that a is odd and the second that a is even (since
k > 0). Thus p. = —I + 0 and we have the following:
(1)
x = -a2 - 4bc + 4c2 + 4ab,
(2)
-2*~'y
= -c2 - ab + b2 + 2ac,
and
(3)
0 = a2 + 4¿»c - t32- 2ac.
By (1), a is odd. From (3) it follows that b is odd and
(4)
2c(a - 2b) = a2 - b2.
Multiplying (2) by 4(a - 2b)2 and using (4) we have
(5)
-2*+1y(a - 2A)2= 3(a - b)(a3 - 3a2b + 3ab2 - 5b3).
Let g = (a, b)>0,A = a/g and B - b/g. Thus gB is odd, A - B > 0 and
(A, B) =\.(A - 2B, A - B)=\, since 2(A - B) - (A - 2B) = A, and (A
- B)-(A - 2B) = B. (5) yields
(6)
-2k+xy(A - 2B)2= 3g2(A - B)(A3 - 3A2B + 3AB2 - 5B3).
Hence 3g2|(^ - 2fi)2. Thus 3g\A - 2B. Therefore (A - B, 3) = 1 and A +
B = A - 2B = 0 (Mod 3). Consequently (A - 2B, A + B) = 3, since (A -
2B) + 2(A + B) = 3A and (A + B) - (A - 2B) = 3B. It follows from (4)
that
2c • (A - 2B)/3g = (A - B)-(A + B)/3.
Hence A - 2B = ±3g. By (6),
(7)
-3 • 2*+ 1y = (A - B)(A3 - 3A2B + 3AB2 - 5B3).
Therefore A — B = 2r, where 0 < r < k + 1. (7) implies that
-3 • 2*+1-'y = (A - B)3- 4B3 = 23r - 4B3.
Thus /-^O.
have
Therefore 3r > 2. Hence k + 1 - r = 2 [3, Lemma 7] and we
(8)
23r"2 + 3y = B3.
If r = 2t, then (23'"1)2 + 3y = B3. There are no solutions in this case by
Table I. Thus r = 2t + 1 and by (8),
(23,+2f+24y = (2B)3.
Using Table I, y = - 1 and (2B, 23,+2>= < - 2, 4> or <10,32>.
If <2B, 23,+2>= < - 2, 4>, then B-1,
t = 0, r = 2t + I = \,A = B +
2r = 1, k = 1 + r = 2 and n = 3k + 1 = 7. Since ±3g = A - 2B = 3 and
g > 0, g = 1, a = g/4 = 1 and b = gß = -1. By (4), (1) and (*), c = 0,
x = -5 and |y| = 1.
If (2B, 23,+2>= <10, 32) we obtain similarly, k - 4, x - 1915 and |y| =
48383.
Proposition
3. // (*) foWs w/iA « = 3k + 2 and xy odd, then
(y,n,x, |y|> = <-l,2,
- 1, 1>or <-l, 8, 11,23>.
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216
STANLEYRABINOWITZ
Proof. If k = 0, then (9y)2 + 108y = (3x)3. The solution of this equation
is given in Table I. Thus we may assume that k > 0.
(x, 3) = 1. By [3, Lemma 5], 3v2 = x3 - 23*+2y= ± 1 ± 4 (Mod 9). Thus
(y, 3) = 1. By (*), 3y2 = aß, where
a = x-2ky92
and ß = x2 + 2ky92x+ (2k92f.
As in Proposition 2, a > 0 and (a, ß)a\a3 = (1 + 9)3(- 1 + 9). Also
x - 2*y = x3 - 23*y s x - 23*+2y = 3y2 = 0
(Mod 3).
Let u = (x - 2*y)/3. Then a = (1 + 9)(x - 2u + (2m - x)9 + u92).
Following the argument of Proposition 2 we see that since N(a) = 3y2,
a/(\ + 9) = p(a + b9 + c92)2,where p = 1 or -1 + 9, and c > 0.
If p = 1, then
x = a2 + 4bc + 2¿?2+ 4ac and 0 = a2 + 4bc + 2c2 + 2ab.
The first equation implies that a is odd and the second that a is even. Hence
p = —1 + 9. Consequently
(9)
x = - a2 - 4bc + 4c2 + 4ab,
(10)
0 = - c2 - ab + b2 + 2ac,
and
(11)
-2ky = a2 + 4bc - b2 - 2ac.
a is odd by (9). Therefore since k > 0, b is odd by (11). (10) implies
(12)
a(b - 2c) = b2-
c2.
Multiplying(11) by -(b - 2c)2and applying (12) yields
(13)
2*y(¿? - 2c)2= 3c(c3 - 6¿?c2+ 6¿?2c- 2Z?3).
Let g = (b, c) > 0, B = b/g
and C = c/g. Therefore gB is odd, C >
0, (B, C) = 1 and by (13),
(14)
2ky(B -2C)2=
3g2C(C3 - 6BC2 + 6B2C - 2B3).
As in Proposition2, (B - 2C, B - C) = 1, 3g\B - 2C, and (B - 2C, B +
C) = 3. Thus (C, 3) = 1. It followsfrom (12) that
a ■(B - 2C)/3g = (B - C)(B+
C)/3.
Therefore B - 2C = ±3g. By (14),
3 • 2*y = C(C3 - 6BC2 + 6B2C - 2B3).
Hence C = 2r, where 0 < r < k. Thus
(15)
3 • 2*"ry = 23r - 3 ■22r+1fi + 3 • 2r+152 - 253.
r¥=0, since k > 0. Therefore 3r > 2r + I > r + I > 2. By (15), k - r = 1
and
(16)
23r~x +3y
= (2r - B)3.
If r = 2t, then (23'+1)2+ 24y = (2(2r - B))3. There are no solutions of this
equation by Table I. Thus r = 2t + 1 and (16) becomes
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v3
THE SOLUTIONOF 3y2 ± 2"n _= X
217
(23'+x) + 3y = (2r - B)3.
Using Table I we find that y = -1, 23,+1 = 2, and 2r - B = 1. Therefore
/ = 0, r = It + 1 = 1, B = 2r - 1 = 1, C = 2' = 2, /c = r + 1 = 2, and
±3g = B - 2C = -3. Hence g = 1, b = gB = 1, c = gC = 2, and by (10),
a = 1. By (9), x = 11 and finally 3y2 = x3 - 23k+2y= 3(23)2.
Theorem. /<// iAc solutions of (*) are g/uew in Table II, vvAerex = 2ge and
y = ±2hf.
Explanation of Table II. 77is given modulo 6 and is nonnegative. If
/ = 0, the value of A is irrelevant. The solutions are numbered for reference in
the proof.
Table II
y n (modulo 6)
-
-
1
1
1
1
1
1
1
1
1
1
1
1
1
3g
0
77
3
77
3
77-3
4
77 + 2
0
77
1
77- 1
1 (n>l)
77-7
1 (n > 13) 77-13
2
77-2
2 (77 > 8)
77 - 8
2
77 + 1
2
77 + 1
3
77
2A
1
1
11 77-3
1
n
f
Solution
number
0
0
21
1
2
1
0
1
1
- 1
1 n- 1
-5
77-7
1915 77-13 48383
- 1 « -2
1
11 77-8
23
1
61
- 1
n
1
n
389
0
3
4
5
6
7
8
9
10
11
12
13
Proof. By direct calculation the above can be shown to be solutions of (*).
Suppose now that (*) holds. Obviously x ^ 0. If y = 0, then x3 —2/ly and
therefore 3|t7. Thus solution 1, 2, 5 or 13 holds.
Assume that y J= 0. Therefore x = 2% and y = 2hf,where efis odd. By (*),
(17)
3 • 22A/2+ 2"y = 23ge3.
It follows from [3, Lemma 7] that
(18)
2A = 3g < n,
(19)
or
2A = 77< 3g,
(20)
3g = 77< 2A.
If (18), then 2A = 3g = 6w. By (17), 3/2 + 2"-6my= e3. Propositions 1, 2
and 3 imply that solution 3, 6, 7, 8, 9 or 10 must hold.
If (19), then n = 6w + 2i, where i = 0, 1 or 2. By (17), (9 • 2J/)2+ 27 • 22iy
= (3 • 2g~2we)3.By Table I, solution 4, 11 or 12 holds.
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218
STANLEY RABINOWITZ
If (20), then n = 6w + 3/', wherey = 0 or 1. By (17),
(9 • 2h-3wff+27 • 23jV= (3 ■2Jef.
There are no solutions by Table I.
Note added in proof. I wish to thank Professor N.
informing me that the results of my doctoral dissertation,
paper and [3] are taken, were also determined by Professor
his doctoral dissertation "Elliptic curves with conductor N =
M. Stephens for
from which this
F. B. Coghlan in
2m3n" completed
in 1967 at Manchester.
References
1. O. Hemer, On the Diophantine equation y2 — k = x3, Thesis, Univ. of Uppsala, Almqvist &
Wiksells,Uppsala, 1952.MR 14, 354.
2. W. J. LeVeque, Topicsin number theory, Vol. II, Addison-Wesley, Reading, Mass., 1961.
3. S. Rabinowitz, 77iesolution ofy2 ± 2" = x3, Proc. Amer. Math. Soc. 62 (1977), 1-6.
4. B. L. van der Waerden, Modern algebra, Vol. I, Ungar, New York, 1949.
Department of Mathematics and Computer Science, Kjngsborough Community College/
CUNY, Brooklyn, New York 11235
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