mathematics of computation
volume 59,number 199
july
1992,pages 195-202
NU-CONFIGURATIONSIN TILING THE SQUARE
ANDREW BREMNER AND RICHARD K. GUY
Abstract. In order to tile the unit square with rational triangles, at least four
triangles are needed. There are four candidate configurations: one is conjectured not to exist; two others are dealt with elsewhere; the fourth is the "nuconfiguration," corresponding to rational points on a quartic surface in affine
3-space. This surface is examined via a pencil of elliptic curves. One rank-3
curve is treated in detail, and rational points are given on 772 curves of the
pencil. Within the range of the search there are roughly equal numbers of odd
and even rank and those of rank 2 or more seem to be at least 0.45 times as numerous as those of rank 0. Symmetrical solutions correspond to rational points
on a curve of rank 1, which exhibits an almost periodic behavior.
1. Introduction
The problem of tiling an integer-sided square with integer-sided triangles
is considered in [6]. This is clearly equivalent to the problem of tiling the
unit square with rational-sided ("rational") triangles. A dissection into two
such triangles is trivially impossible, and it is known (see [6] for a proof and
references) that a dissection into three rational triangles is also impossible.
Accordingly, the first interesting case is a dissection into four rational triangles. Figure 1 shows the four candidates for such a configuration, named respectively chi-, delta-, kappa-, and nu-configurations. A rational chi-configuration is
conjectured to be impossible. A detailed discussion of how to obtain "all" deltaand kappa-configurations is given in [1]. Here we give a parallel discussion of
nu-configurations.
In Figure l(v), take OB, OA as axes and U the point (1, 1). If P is the
point (X, 1) and Q the point (X + Y, 0), then the rationality.of the lengths
OP, PQ, QU immediately implies that
X2 + 1 = D,
(1)
72+l=D,
(1 -X -Y)2+
1 =D.
We call a rational number, x, satisfying x2 + 1 = D, a rectangular number.
Then it is clear from ( 1) that nu-configurations are characterized by solutions
of the equation
(2)
X + Y+ Z = l
Received by the editor November 26, 1990.
1991MathematicsSubjectClassification.Primary 11D25, 11G05, 11Y50.
© 1992 American Mathematical Society
0025-5718/92 $1.00+ $.25 per page
195
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196
ANDREW BREMNER AND R. K. GUY
_
_
A
-P
O
(X)
(A)
(K)
U
Q
»
iv)
Figure 1. Four candidates for a tiling of the square
with four rational triangles
in which X, Y, Z are rectangular numbers. We investigate (2) in the manner
of Bremner and Guy [1].
2. The parameters
(r,s),
(u, v), (m, n)
In (2), there is no loss of generality in putting
y _ r2 - s2
2rs
_ u2 - v2
'
2uv
_ m2 - n2
'
2mn
with r ± s (r prime to s ), u J_ v, and m ± n, so that we seek integer
solutions m, n, r, s, u, v to the equation
(3)
mnuv(r2 - s2) + mnrs(u2 - v2) + rsuv(m2 - 2mn - n2) = 0.
Geometrically, (3) represents a quartic surface in affine three-dimensional space
with coordinates m/n, r/s, u/v . It seems an unlikely hope that there is
any straightforward description of all the rational points on this surface. In
order to obtain results, we shall assume, much as in [1], that the ratio m/n is
predetermined, so that (3) may be considered as an elliptic curve over Q(m/n).
3. Symmetrical
nu's
The special case where r/s = m/n , so that Figure l(^) is symmetric under
rotation through n, is considered in §7 of [6], so we merely summarize the
results. The solutions correspond to rational points on the curve
(4)
T2 = a(a2 + 6o + 4)
which has rational rank 1, with a generator P = (-1, 1) of infinite order. The
solutions (r: s; u: v) corresponding to P, 2P, 3P, and 4P are (1: 0; 0: 1),
(-3: 4; 3: 2), (75: 52; 50: 39), and (425: - 5928; 221: 5700). The first is
degenerate, the second and third correspond to the proper tilings of Figures 2
and 3, and the fourth to the improper tiling of Figure 4.
All symmetrical nu-configurations are given by a single infinity of solutions
satisfying the recurrence relations
rn+x: sn+i = r„(2(r„ -sn)un
u„+x: vn+x =vn(2(rn
-s„)un
-s„v„):
-sn(2rnu„
-s„v„),
-snv„):
un(2rnu„ -s„v„).
It was observed in Table 7.1 of [6] that these solutions are almost periodic,
with a period close to 282. This is explained as follows. It is well known that
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NU-CONFIGURATIONSIN TILING THE SQUARE
Figure 2. n = 2
197
Figure 3. n - 3
Figure 4. n-4
an elliptic curve over C is isomorphic to the quotient of C by some lattice A,
the isomorphism being given by the Weierstrass p-function, ( p(z), p'(z)) •-►z
mod A ; and the points of the curve over C are in one-to-one correspondence
with the points of the fundamental parallelogram of the lattice (Figure 5), where
OJt and a>2 are known as the periods of the curve.
For the curve (4), then, O corresponds to the point at infinity o, and œ2
is actually real. Further, as a point Q traverses the right-hand branch of the
curve (containing (0, 0) ), then the associated parameter z of the p-function
traverses the edge of the fundamental parallelogram from O to co2 (containing
\œ2 ). It follows that the associated parameter z of Q is of type aco2, 0 <
a < 1 . Suppose now that a/b is a good rational approximation to the real
number a. Then Q will display a 'period' of b, in the sense that bQ is close
to o. But a is simply determined as the ratio of the integrals
r°° dx_ I
r°° dx_
Jxqy I k y
(the denominator here being the real period co2).
O
>2
Figure 5. Fundamental
with periods u>x, a>2
to:
parallelogram
of the lattice
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198
ANDREW BREMNER AND R. K. GUY
Performing the integration (kindly computed by John Hebron) for the point
Q = 2P = (9/4, 57/8), we find
a = 0.2411328840030960787192512....
In order to find good rational approximations to a, compute the continued
fraction. The partial quotients are
4,6, 1,3, 1,26,7,2,
1,2, 1,62,9,4,5,
1, 1,50,2,2,....
The large partial quotients 6,26,62,50,...
ensure that the denominators of
the convergents obtained by truncating them, namely
1 _34_ 74165 1980618265
4' 141' 307569' 8213804074'
give very good 'periods' of 2P. It follows that 4, 282, 615138, 8213804074
give very good estimates for a 'period' of P.
It is not clear to us whether the large partial quotients in the above continued
fraction are simply a curiosity. For a continued fraction without large partial
quotients, there would be no obvious 'period' of small size. A similar example
is worked by Don Zagier in Method 3 of [12].
4. Fixing (m,n)
leads to an elliptic
curve
In order to find solutions containing a given slope, (m, n), write equation
(3) as a quadratic in r and s :
■y „ (u2-v2
r2 + 2rs —-+-
m2 -2mn
\ 2uv
- n2\
2mn
J
-,
- s2 = 0,
and r/s will be rational just if the discriminant is square:
u2 - v2
m2 - 2mn - n2^
V 2mv
2m«
J
Write u/v = U and
m2-2mn-n2
-~-=
(6
k,
2mn
so that there is a birational map between (5) and the elliptic curve
(7)
E:t2
= o(o2 + (k2 + 2)o+1)
given by
a = Ua+2-(U2 + 2KUwith inverse
X - KO
1),
t = U(a + 1) + ko,
T + KO
K
U =-—
,
a
a + 1 '
2cr
a +V
There is a 2-isogeny, v', between the curve (7) and the curve
(8)
E':T2
= S(S - k2)(S - (k2 + 4))
given by
(9)
(S,T) = u'(a,T)=(^2,{l-^j^
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NU-CONFIGURATIONS IN TILING THE SQUARE
199
The corresponding isogeny v: E' —>E, where uv1 is multiplication
E, is given by
(10)
(a,x) = viS,T)=(^,\
k2(k2 + 4)
S2
by 2 on
I
For terminology and details of the calculation of the rank of E (and E' ) over
the field k = Q(m/n), we refer the reader to Bremner and Guy [1]. Here we
simply compute the groups G* and G*, of ^-covers of E, and i/'-covers of
E', which actually possess points defined over k . Then the rank g of E can
be determined from the relation
2g+l = [G:][G*,]/2.
Consider first the curve (7), which, by putting X = m/n in (6), we may write
in the form
(11)
T2 = a(a2 + (A4- 4A3+ 10A2+ 4X+ \)a + 16A4).
The ^-covers of (11) are obtained in the classical manner by putting a =
Sa2/b2, with 8, a, b e Z[X], 8 squarefree, a Lb . This gives
(12)
S2a4 + (Xa- 4X3+ 10A2+ 4X+ l)ôa2b2 + l6X4b4 = ôc2.
The elements of the group of v -covers of E are in one-one correspondence with
the curves (12). For such a cover to be locally solvable, it is clearly necessary
that 8 divide 16X4, and ô squarefree forces the only possibilities for ô to be
(13)
S = ±l,±2,±X,±2X.
Rewrite (12) as
[20a2 + (X4- 4X3+ 1OX2+ 4X+ l)b2]2
- (X4- 4X3+ 18A2+ 4X+ l)(X2 -2k-
l)2b4 = 48c2.
Suppose that 8 is a quadratic nonresidue modulo the polynomial
PÍX) = X4- 4X3+ 18A2+ 4A+ 1.
Then (14) forces
28a2 - SX2b2= c = 0 mod piX).
In particular, 8a2 = 4X2b2, so that a = b = 0 mod /?(A), a contradiction.
Accordingly, for local solvability, 8 must be a quadratic residue modulo
p(X).
Let 6 denote a root of p(X). Then 8 is a square
mod p(X) if and only if
8(6) is a square in Q(d).
Take 6 = l+2/+v/-2 + 4i. We claim that the possibilities ô = ±2, ±X, ±2X
from the list (13) contradict the above local criterion. Since -1 belongs to
Q(6)2, it suffices to show that 2,6, and 26 do not. Consider, in Q(6), the
first-degree prime ideal factor p5 of the principal ideal generated by 5, for which
i = 2 and 0=1. If either 2 or 20 were a square in Q(0), it would now follow
that 2 is a square mod ps, which is impossible.
Similarly, consider the first-degree prime ideal Pn(0) of the ideal generated
by 13, for which i = 8 and 0 = 2. If 0 were a square in Q(0), it would imply
that 2 is a square mod pu , which is also impossible.
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200
ANDREW BREMNERAND R. K. GUY
We conclude that the only possibilities for 8 in the list (13) are 8 = ±1 ;
and there exist respective global points given by (a, b, c) = (1,0,1)
and
(2X, 1, 2A(A2- 2X- 1)). Hence [G*u]= 21 .
In a similar way, consider the curve (8) in the form
(15)
T2 = S(S - (X2-2X-
l)2)(S - (X4- 4X3+ IU2 + 4X+ 1))
and put S = AA2/B2, with A, A, B e Q[X], A squarefree, ALB.
(16)
(A^2 - (X2-2X-
This gives
l)2B2)(AA2 - (X4- 4X3+ ISX2+ 4X+ l)B2) = AC2,
and, for local solvability of (16), A must divide
(X2-21-
\)2(X4- 4X3+ 18X2+ 4X+ 1) ;
it is also clear that both the leading coefficient and the constant term in A must
be positive.
Hence, there are just the possibilities
A=l
and
A = X4- 4X3+ 18A2+ 4X+ 1,
with respective global points (A, B, C) = (1, 0, 1) and (1, -1, 0). So [G*,]
= 2'.
Finally, the rank g satisfies
2*+2 = [g:,][g¿]
and accordingly g = 0. So numerical instances of the nu-configuration may
be less common than the delta-configurations of Bremner and Guy [1]. But we
shall see in §10 (in the Supplement at the end of this issue) that, as often as
not, the ranks for specific slopes are positive. And in the next section we find
an infinite family of parametrized nu-solutions.
5. Families of curves of rank one
Take the ratios m/n , r/s,
u/v to satisfy n = s —v . Then (3) becomes
n2(mr + ru + urn) + 2nmru - mru(m + r + u) —0
and accordingly, a solution is furnished by setting mr + ru + urn = 0. There
results
u--j
mr
m2 + mr + r2
m+r
n = —^-;—>
2(m + r)
leading to the parametrization
m = 2p(p+ 1)
r = 20+1)
n
s
p2 + p+l'
p2 + p+l'
,
u_
v
-2p
. p2 + p+l'
From this solution we may generate infinitely many further solutions in the
standard manner, as follows. Set
m _ 2pip + 1)
n
p2 + p+ 1
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nu-configurations
in tiling the square
201
and work over the field Q(p). From (6), k now takes the form
p4 + 2p3 + 7p2 + 6p+l
4p(p + l)(p2 + p + 1)
From the values (17) and the maps to the curve (7) we see that the curve
possesses the Q(p)-point
hrï
(18)
r\
( 4^+1)
PP-tn
= {a>r)={p2
+ p+l)2>-(P2 (2p+l)(p4
+ 2p3
+ 7p2 + 6p+iy
+ P+D3-
We have verified that the rank of E defined over <Q(p) is equal to one, and it
seems plausible that the point P at (18) is indeed a generator for the group of
points of E defined over Q(p), although this has not been specifically checked.
It is, however, easy to see that P is of infinite order, and accordingly, the
points NP, N 6 N, furnish an infinite family of examples of parametrized
nu-solutions. The case N = 2, for example, leads to the solution
u=
(p + l)(p2 + p-
l)2(p4 + 2/i3 + lp2 + 2p + 1)
v ~ 2p(2p + l)(p2 + p+ l)(p4 + lp? + lp2 + lOp + 5) '
r = 2(p+l)(2p+l)(p2
+ p+l)(p4 + 2p3 + 7p2 + 2p+l)
s
p(p2 + p-l)2(p* + 2pi + 7p2+l0p + 5)
Another infinite family of parametrized nu-solutions is given by
m
n
p2 + 1
p(p2 - 2p + 2) '
r_p(p2-2p-2)
5
2p2 -2p-\
u
v
p(p2 + I)
2p2 - 2p - 1
and may be handled in a similar manner.
Sections 6-10 of this paper appear in the Supplement at the end of this issue.
Bibliography
1. A. Bremner and Richard K. Guy, The delta-lambda configurations in tiling the square, J.
Number Theory 32 (1989), 263-280.
2. J. P. Buhler, B. H. Gross, and D. B. Zagier, On the conjecture of Birch and Swinnerton-Dyer
for an elliptic curve of rank 3, Math. Comp. 44 (1985), 473-481.
3. J. W. S. Cassels, Diophantine equations with special reference to elliptic curves, J. London
Math. Soc. 41 (1966), 193-291.
4. David A. Cox and Steven M. Zucker, Intersection numbers of sections of elliptic curves,
Invent. Math. 53 (1979), 1-44.
5. Gerd Faltings, Endlichkeitssätze für abelsche Varietäten über Zahlkörpern, Invent. Math. 73
(1983), 349-366; erratum 75 (1984), 381.
6. Richard K. Guy, Tiling the square with rational triangles, Number Theory and Applications
(R. A. Mollin, ed.), NATO Adv. Study Inst. Ser. C: Math. Phys. Sei., vol. 265, Reidel,
Dordrecht, 1989, pp. 45-101.
7. Barry Mazur, Modular curves and the Eisenstein ideal, Inst. Hautes Études Sei. Publ. Math.
47(1977), 33-186.
8. _,
Rational isogenies of prime degree (with an appendix by Dorian Goldfeld), Invent.
Math. 44(1978), 129-162.
9. Joseph H. Silverman, Computing heights on elliptic curves, Math. Comp. 51 (1988), 339-
358.
10. Larry C. Washington, Number fields and elliptic curves, Number Theory and Applications
(R. A. Mollin, ed.), NATO Adv. Study Inst. Ser. C: Math. Phys. Sei., vol. 265, Reidel,
Dordrecht, 1989, pp. 245-278.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
202
ANDREW BREMNERAND R. K. GUY
H.A.
Weil, Sur les courbes algébriques et les variétés qui s'en déduisent, Hermann, Paris, 1948.
12. Don Zagier, Large integral points on elliptic curves, Math. Comp. 48 (1987), 425-436;
Addendum, ibid. 51 (1988), 375.
Department
of Mathematics,
Department
of Mathematics
Alberta T2N 1N4, Canada
Arizona State University,
and Statistics,
Tempe, Arizona 85287-1804
The University
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of Calgary,
Calgary,
mathematics of computation
volume 59.number 199
july 1992,pages si-s20
Supplement to
NU-CONFIGURATIONSIN TILING THE SQUARE
ANDREW BREMNERAND RICHARD K. GUY
6. Lemmas
restricting
the
rank
in specific
cases
Although the generic rank of (7) is zero, the rank of the specialized curve, when m
and n are given numerical values, is often positive. We obtain arithmetic information
about the rank in the manner of §4 of [1]. To make the dependence on m,n more
explicit, we introduce some notation and note some congruences which will be useful
in the sequel. Write
K --=m2 - 2ran - n2 = (m - nf - 2n2,
L = K2 + 8mV
= m* - 4m3n + lfjmV
M = K2 + 16mV
so that k = A'2/4m2n2,
+ 4mn3 + n\
= m" - 4m3n + 18m2n2 + 4mn3 + n",
k2 + 2 = L/4m2n2,
k2 + 4 = M/4m2n2.
We may take m ± n,
so that K L mn, L X ran, M J_ mn, and
Fact
(a) If m,n are of opposite parity, then K s ±1 mod 8, and each of K2, L, and
M is congruent to 1 mod 16.
Fact (b) If m,n are both odd, then K = 2 mod 4, and A'2 = 4, L = 12, and M =
20 mod 32.
Fact
Fact
(c)
Each odd prime factor of K is congruent
to ±1 mod 8.
(d)
Each odd prime factor of M (the sum of two squares)
is congruent
to 1 mod
4.
We also change the scale, replacing
(4m2n2cr,8m3n3r)
by (c, r), so that
(7), which
is
64m6n6T2 = 4mV<r(16m4nV
+ L ■4m2n2a + 16mV),
becomes
r2 = <t(<t2+ La + 16mV);
(19)
and replace (4m2n25,8m3n3T) by (S,T), so that (8), namely
64m6n6T2 = 4mV5(4m2n25
- A-2)(4m2n2S - M),
becomes
T2 = S(S - K2)(S - M).
(20)
In (19) we substitute a = ba2j\?,T = Sac/b3, with 6,a,b,c G Z,<5 squarefree, a _L6
and a, b, c > 0, and obtain
Sc2 = ¿V + LSa2b2+ 16mV64,
(21)
© 1992 American Mathematical Society
0025-5718/92$ 1.00+ $.25per page
SI
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