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MATHEMATICS
OF COMPUTATION
VOLUME 42, NUMBER
165
JANUARY
1984, PAGES 257-264
On the Equation Y2 = X{ X2 + p)
By A. Bremner and J. W. S. Cassels
Abstract. Generators are found for the group of rational points on the title curve for all primes
p = 5 (mod 8) less than 1,000. The rank is always 1 in accordance with conjectures of Seltner
and Mordell. Some of the generators are rather large.
1. Let p be a positive prime,
(1)
/7 = 5(mod8).
It is easy to see that the Mordell-Weil rank of
(2)
Y2 = X(X2 + p)
is at most 1 (e.g. Section 5 of Birch and Swinnerton-Dyer [1]); and the Selmer
conjecture [5] predicts rank exactly 1. As we shall note below, this is equivalent to a
conjecture of Mordell [3], [4].
We shall verify this conjecture for all p < 1,000. Table 1 gives for each p a point
P0 which, together with the point (0,0) of order 2, generates the entire Mordell-Weil
group. Some of the generators P0 are rather large, the most startling being that for
p = 877, namely
3
_ 37 5494528127162193105504069942092792346201
621598777687 1505425463220780697238044100 '
2562562679889268093887768340455 13089648669153204356603464786949
490078023219787588959802933995928925096061616470779979261000 '
A rational point (X,Y)on
(4)
for integers R, S, T with
(3) is of the shape
X=R/S2,
(5)
R>0,
The height H(X, Y) is by definition
(6)
Y=T/S3
(R,S)=l.
H(X,Y) = max(R,S2),
so that the height of (3) is - 3.75 x 1041.It is of interest to compare this with the
discriminant |A| = 4p3 of (2), which for p = 877 is - 2.70 x IO9.Hence H(P0) |A|441.Lang observed to us that in tables of elliptic curves and generators published
to date the heights of generators are never much greater than |A|2. This, in our view,
is almost certainly because these tables cover only curves with relatively small
ReceivedDecember 28, 1982.
1980 MathematicsSubject Classification.Primary 10B10, 14K07.
©1984 American Mathematical Society
0025-5718/84 $1.00 + $.25 per page
257
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258
A BREMNER ANDJ. W. S.CASSELS
discriminant. Further, unless one has strong reason to believe that a rational point
exists, there is a marked reluctance to persevere in a search once the numbers cease
to be small.
2. In (2) clearly either X or pX is square. Since
(7)
(X,Y) + (0,0) = (p/X,-pY/X2).
we may suppose that X is a square. Then
(8)
Y = rt/s 1.3
X=r2/s2,
for integers r, s, t with r,s coprime and
(9)
r4+ps4
= t2.
It was for this equation that Mordell made his conjecture mentioned above.
Clearly
(10)
rmO,
t*0(modp),
and (1) implies that
(11)
r = t= l,
s = 0(mod2).
We choose the sign of t so that t = 1 (mod4), and then
(12)
/sl(mod8)
by (9) and (11). Since (9) can be written as
(13)
(t + r2)(t - r2) = ps\
we have t + r2 = 2a4 or 2pa4 for some odd a. The first alternative leads to a point
Q on (2) with (X,Y)= 2Q. Hence by the "infinite descent" argument we may
suppose that there is a coprime pair of integers a, b with
( 14)
r2 = pa4 - 4b4.
a = b=l
(mod2)
and
(15)
s = 2ab,
t=pa4-r4b4.
We can write ( 14) in the shape
(16)
(r + 2ib2)(r-
2ib2) = pa4
and consider factorization in Z[i]. By (1)
(17)
p = u2 + 4v2,
where u, v are odd and without loss of generality
(18)
v= 1 (mod 4).
The sign of r may be chosen so that r + 2ib2 is divisible by u + 2iv, and then (16)
implies
(19)
r + lib1 = (u + 2iv) • unit ■(c + id)4
for some c, d with
(20)
c2 + d2 = a.
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ON THE EQUATION Y2 = X( X2 + p )
259
On considering (19) modulo 8 and using (18), we find that the unit is necessarily 1.
Hence on equating real and imaginary parts,
(21)
b2 = u(/2 - m2) + ulm,
with
(22)
l = c2-d2,
m = 2cd.
If there are solutions of (2), then (21) must have rational solutions. This turns out
to be the case for all the/7 under consideration, and so every solution of (21) is given
by one of a finite number of parametrizations
(23)
/ = fi(M).
m = q2(6,x¡,), b = q,(6,xp).
Here qx, q2, <73are known quadratic forms with rational integer coefficients and 0, xp
are integers to be found. It turns out that only one parametrization is compatible
with the other conditions.
We have now
(24)
qx(6,xp) + iq2(6,xp) = (c + id)2.
Arguing as before, but in Z[i], we have
(25)
6 = Qx(X,p), xp= Q2(X,p), c + id=Q3(X,p),
where Qx, Q2, Q3 are known quadratic forms with coefficients in Z[i] and where X, p
are elements of Z[i] to be found. Again, only one parametrization turns out to be
compatible with the other conditions. The condition that 6, xpare real leads to a pair
of simultaneous homogeneous quadratic equations in the four variables Re\, Im\,
Re/i, Imp. These were the equations searched for solutions, though most of the
entries in Table 1 could be spotted at an earlier stage in the process. The primes 317,
797, 877, 997 required an HP67, but the other solutions were found by hand.
At the suggestion of the referee we illustrate the last part of the argument and take
(26)
p - 877, u = 29, v-3.
Then (21) is
(27)
b2 + 3/2 - 3w2 - 29/m = 0.
The left-hand side vanishes for (b,l,m) = (7,2,1) and so, by a standard algorithm,
(27) is equivalent to
(28)
-LM + 877/V2= 0,
where the forms
(29,)
L= 14/3- 17/-64m,
(292)
M = 50746-6242/-
(293)
N = 9b-lll-4\m
23035m,
are unimodular. On taking -b for b if need be, we have 877 + M, so (28) implies
(30)
±L = 87702, ±M = <b2, ±N = 6tt>
for some integers 6, (/>and some choice of sign.
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260
A. BREMNER AND J. W. S. CASSELS
On solving for b, l, m and putting
(31)
<t>
= -t^ + 5630,
we obtain
(32,)
±b=
-7xp2 - llxp6 + 2162,
(322)
±/=
-2f- + 6xpe- 302,
(323)
±m=
-xp2 -4xpe - IO2.
The lower sign is incompatible 2-adically with (22), so we must take the upper sign.
By (22) we have
(33)
/ + /« = (c + /¿)2 = Y2(say);
and so
(34)
y2 + (2 4- i)xp2 + (-6 + 4i)xpe + (3 + li)02 = 0
by (322), (323). This is equivalent to
(35)
(6 -29i)S2 + RT=0,
where
(36,)
S = (l + i)y + (-4 + i)0-
ixp,
(362)
R = (3 + 2/)y + (-10 + 5i)0-2ixp,
(363)
7= (-15 + 6i)y + (8-45i)«
+ (14 + 4i)^.
Hence
(37,)
20 = (-4 - 3i)R + (22 + 10/)5 + iT,
(372)
2xp= (-l
(373)
2y = (-15 + 6i)Ä + (70 - 46/)S + (3 + 2i)T.
+2i)R+
(6-
\6i)S + T,
By (35), on taking - y for y if need be, there are Gaussian integers X, p such that
(38)
R = (l±i)X2,
r=(6-29/)(l
±i)p2,
S = (1+/)a/i,
for either the upper or the lower signs. Put
(39)
X = x + iy,
p = u + iv,
where x, y, u, v are rational integers. On substituting (38), (39) in (37, ), (372) we get
(40,)
21?= Fx(x,y,u,v)
+ iF2(x,y,u,v),
(402)
2t/>= Gx(x,y,u,v)
+ iG2(x,y,u,v),
where Fx, F2,GX,G2 are quadratic forms with rational integer coefficients. [There is
a set of forms for each choice of sign in (38).] Hence
(41)
F2(x,y,u,v)
= G2(x,y,u,v)
= 0.
If the lower signs hold in (39), the simultaneous equations (41) turn out to be
2-adically incompatible, so we must take the upper signs. A search yields
(42)
X = 324 - 385L
p = 136 + 145/.
It may be noted that (41) implies congruence conditions on x, y, u, v to various small
moduli, and these greatly facilitate the search.
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ON THE EQUATION Y2 = X( X2 + p )
261
3. Having indicated how the rational points P0 were obtained, we must now show
that they are generators. This requires consideration of heights.
Let (4) be a rational point P on (2). The A'-coordinate of 2/* is
(43)
Xx^{Rl-pS*)2452r2
(*2-'S4)2
4RS2(R2 + pS4)'
We consider only points of the type (8), (9), so p \ R. It is then easy to see that the
numerator and denominator in (43) are coprime. Hence
(44)
//(2/>) = max{(/v2-/>S4)2,
4RS2(R2 + pS4)}.
We distinguish three cases:
(i) 0 « S2/R < 1/2px>/2.Then
H(2P) >(R2(ii)l/2^'/2<
pS4)2 > (9/16)*4 = (9/l6)H(P)\
S2/R*i l.Then
H(2P) > 4RS2(R2 + pS4) > {2/px/2)R4 * {2/px/2)H(P)\
(iii) 1 < S2/R. Then
H(2P) > (R2 - pS4)2 >(p-
1)V
= (p - l)2/7(/>)4.
Hence in any case
(45)
H(2P)>(2/px/2)H(P)\
.
Similarly, but more simply,
(46)
"',
".
H(2P) < p2H(P)\
With the usual notation h(P) = logH(P) it follows that
(47)
4h(P) - ilog(/7/4) < h(2P) < 4h(P) + 21ogp.
Now (43) is a perfect square, so the argument applies to 2P instead of P. By
induction
4"h(P)-
(l/6)(4" - l)log(/>/4) < h(2"P) . .
' '
< 4"h(P) + (2/3)(4" - l)log/7
for every n > 1. The Täte height is
(48)
h(P) = Urn 4-"rj(2nP),
,
n-"x>
SO
(49)
h(P) - (l/6)log(/>/4) « h(P) < h(P) + (2/3)log/7.
Let P0 be one of the points on (2) listed in Table 1. The descent argument shows
that neither P0 nor P0 + (0,0) is divisible by 2. Suppose that P0 is divisible by 3, say
P0 = 3Q. Let q be a prime distinct from 2, p. Then (2) has good reduction modulo q
and'the reduced points P0, Q modq would satisfy P0- 3Q. For each j?0Table 1
gives a prime q such that P0 is not divisible by 3 in the group of points on (2) over
the finite field Fq. Hence if PQ is not a generator we have PQ= kQ for some odd
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264
A. BREMNERAND J. W. S. CASSELS
Ac3s 5 and some rational point Q. By the quadratic property of heights
(50)
h(Po) = k2h(Q)
(e.g. Cassels[2, p. 262]).From this and (49) it followsthat
Mo) < {\/k2)h(P0) + (2/3k2)logp + (l/6)log(p/4)
< (1/25)A(>?0)+ (2/75)log/> + (l/6)log(/>/4).
Even in the extreme case/? = 877 this implies H(Q) < 136, which in turn implies a
solution of (9) with 0 < r, s < 12; a contradiction.
Department of Pure Mathematics
and Mathematical Statistics
Universityof Cambridge
CambridgeCB2 ISB, England
1. B. J. BIRCH& H. P. F. Swinnerton - Dyer, "Notes on elliptic curves II." J. Reine Angew. Math., v.
218, 1965,pp. 79-108.
2. J. W. S. Cassels. "Diophantine equations with special reference to elliptic curves." J. London Math.
Soc.,v. 41. 1966,pp. 193-291.
3. L. J. MORDELL,
"The diophantine equation ,x4 + mv* = :2." Quart. J. Math. (2). v. 18, 1967.pp.
1-6.
4. L. J. Mordell,
y = Dx* + I, Number Theory Colloquium, János Bolyai Math. Soc., Debrecen,
1968.pp. 141-145(North-Holland,Amsterdam,1970).
5. E. Selmer, "A conjecture concerning rational points on cubic curves," Math. Scand., v. 2, 1954,pp.
49-54.
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