56 Normat 61:2, 56–66 (2013)
The Taxicab number 1729
Ulf Persson
Matematiska Institutionen
Chalmers Tekniska Högskola och
Göteborgs Universitet
ulfp@chalmers.se
The story of Hardy visiting Ramanujam at his sick-bed and referring to the boring
number - 1729, on the cab that brought him there, only to be contradicted by the
patient who informed him that it was the smallest integer that could be written
as the sum of two cubes in two different ways (12
3
+1
3
= 10
3
+9
3
) is of course
well-known. Hardy was taken aback and asked about the corresponding number
for fourth powers, only to be told that Ramanujam did not know but suspected it
was very high.
The moral of the story is of course to indicate the amazing familiarity Rama-
nujam had with individual numbers. As a mathematical problem, Hardy probably
did not think it was many cuts above the kind of trivial facts that he ridiculed in
his Apology and which formed the back-bone of such publications as that of Rose
on recreational mathematics (later to be revised by Coxeter). Trivial or not, this is
the kind of thing that can intrigue the young individual, and most of us probably
came across the story early in life.
It is very unlikely that Ramanujam ever systematically looked for such numbers,
his knowledge of the fact was probably just a fortuitous spin-off from other consi-
derations, and even if he had done so, it is unlikely that he would have gotten far.
Nowadays with the advent of high-powered computers anyone with some minimal
skill of programming can make a search in a few seconds. As a curiosity I list all the
primitive solutions up to one million. By a primitive solution I mean one in which
the numbers involved have no common factor, if they do, they arise in a trivial way
from a solution in which there is no common factor but the trivial one.
1729 = 12
3
+1
3
=10
3
+9
3
4104 = 16
3
+2
3
=15
3
+9
3
20683 = 27
3
+10
3
=24
3
+19
3
39312 = 34
3
+2
3
=33
3
+15
3
40033 = 34
3
+9
3
=33
3
+16
3
64232 = 39
3
+17
3
=36
3
+26
3
65728 = 40
3
+12
3
=33
3
+31
3
134379 = 51
3
+12
3
=43
3
+38
3
149389 = 53
3
+8
3
=50
3
+29
3
171288 = 55
3
+17
3
=54
3
+24
3
195841 = 58
3
+9
3
=57
3
+22
3
216027 = 60
3
+3
3
=59
3
+22
3
327763 = 67
3
+30
3
=58
3
+51
3
402597 = 69
3
+42
3
=61
3
+56
3
439101 = 76
3
+5
3
=69
3
+48
3
443889 = 76
3
+17
3
=73
3
+38
3
515375 = 80
3
+15
3
=71
3
+54
3
684019 = 82
3
+51
3
=75
3
+64
3
704977 = 89
3
+2
3
=86
3
+41
3
805688 = 93
3
+11
3
=92
3
+30
3
842751 = 94
3
+23
3
=84
3
+63
3
920673 = 97
3
+20
3
=96
3
+33
3
955016 = 98
3
+24
3
=89
3
+63
3
984067 = 98
3
+35
3
=92
3
+59
3
994688 = 99
3
+29
3
=92
3
+60
3
Incidentally there is no number below a million that can be written as a cube
in three different ways.
While we are at it, we can also try to look for numbers that can be written as
the sum of two fourth powers in two different ways. None is found below a million,
Normat 2/2013 Ulf Persson 57
and indeed Ramanujam was right. Yet a wider, much more time-consuming search
reveals
239
4
+7
4
= 227
4
+ 157
4
158
4
+ 59
4
= 134
4
+ 133
4
542
4
+ 103
4
= 514
4
+ 359
4
292
4
+ 193
4
= 257
4
+ 256
4
631
4
+ 222
4
= 558
4
+ 503
4
502
4
+ 271
4
= 497
4
+ 298
4
The curious thing is that the problem of representing numbers that can be
written as the sum of two cubes in two different ways can actually be completely
solved, at least in the sense that a parametric solution can be written down. This
goes back to the fact discovered early in the 19th century that the cubic surface in
P
3
has twenty seven lines, which can be used to show that it is rational, i.e. that
it has a rational parametrization. In fact it can be parameterized by four cubic
ternary forms. The geometric construction that lies behind it can also be used to
show that this can be done over various fields.
As a warmup we will consider the far more elementary problem of when a number
can be written as a sum of two squares in two different ways.
X
2
+ Y
2
= Z
2
+ W
2
This is indeed a far more elementary problem. In fact we can rewrite it as X
2
≠
Z
2
= W
2
≠ Y
2
or (X + Z)(X ≠ Z)=(W + Y )(W ≠ Y ). This gives a clear clue
how to generate non-trivial solutions. Take a number N which has two different
factorizations u
1
v
1
= u
2
v
2
. If they are of the same parity we can solve for integral
values X + Z = u1,X ≠Z = v1,W + Y = u
2
,W ≠Y = v
2
leading to
(u
1
+ v
1
)
2
4
+
(u
2
≠ v
2
)
2
4
=
(u
1
≠ v
1
)
2
4
+
(u
2
+ v
2
)
2
4
The simplest case with u
1
=5,v
1
=3,u
2
= 15.v
2
=1gives rise to the identity
7
2
+4
2
=1
2
+8
2
= 65 in fact this is the smallest number which can be written as
a sum of two non-zero squares in two different ways. (Otherwise we have of course
5
2
=4
2
+3
2
).
We can think of the parameter space as given by (u
1
,v
1
,u
2
,v
2
) sub ject to the
condition u
1
v
1
= u
2
v
2
. This condition is given by another quadric in P
3
, so in fact
it is rather tautological. The difference being that we are more adept at finding two
numbers that multiply to the same pro duct, then finding two different differences
of squares being equal, although the two things are equivalent.
We can also use geom etry to get a parametrization. The quadric X
2
+ Y
2
=
Z
2
+ W
2
has an obvious solution P =(1, 0, 1, 0). As our parameter space we
consider points on the hyperplane X =0i.e. points (0,u,v,w). The lines joining
them to P have parametric representation
⁄(1, 0, 1, 0) + µ(0,u,v,w)=(⁄, µu, ⁄ + µv, µw)
inserting into the quadric gives the binary quadric
⁄
2
+ µ
2
u
3
≠ (⁄ + µv)
2
≠ µ
2
w
2
58 Ulf Persson Normat 2/2013
which simplifies to
µ(≠⁄(2v)+µ(u
2
≠ v
2
≠ w
2
))
This will of course vanish for µ =0corresponding to the point P . The residual
intersection with the quadric will be given when we set the other factor to be zero,
i.e.
⁄ = u
2
≠ v
2
≠ w
2
µ =2v
Plugging that in in the parametric representation of the line we get the para-
metrization of the quadric given by
(u
2
≠ v
2
≠ w
2
, 2uv, u
2
+ v
2
≠ w
2
, 2vw)
and the reader can easily verify the polynomial identity
(u
2
≠ v
2
≠ w
2
)
2
+(2uv)
2
=(u
2
+ v
2
≠ w
2
)
2
+(2vw)
2
Numbers which are the sum of two squares or two cubes
Given a number n one can verify that it is the sum of two squares without having to
find them. It is well-known that this happens iff every prime-factor of type 4k +3
has even multiplicity in the prime-decomposition of n. The number of different
representations is given by 2
k≠1
where k is the number of prime-factors of type
4k +1. If there are none, the number of representations is 1.
The key idea is that a prime p of type 4k +1 in Z splits into two non-associated
primes a ± ib in the ring Z[i] of Gaussian integers, while 2=(1+i)
2
becomes a
square. For such primes p we have p = Nm(a+ib)=a
2
+b
2
and the representation
is unique (up to trivial m odifications).
In particular if there are many prime-factors of the right type, there will be
many representations as a sum of two squares. E.g. we have for 1105 = 5 ◊13 ◊17
we have
9
2
+ 32
2
= 23
3
+ 24
2
= 33
2
+4
2
= 31
2
+ 12
2
This comes from 5=(2+i)(2≠i), 13 = (3+ 2i)(3 ≠2i), 17 = (4+ i)(4≠i), allowing
us to pick up four different pairs of conjugate Gaussian numbers which multiply to
1105.
Now to find the representation of a prime p of the right type as a sum of square
can of course be done by trial and error for small primes, or by computer searches for
somewhat larger primes. One way of speeding up the process is to find a numb e r
n such that p|(n
2
+ 1) which is equivalent to solving the congruence equation
x
2
= ≠1(p).
This can be done by taking powers of elements mo dulo p,sayifp =37then consi-
der 2, 4, 8, 16, 32, 27, 17, 34, 31, 25, 13, 26, 15, 30, 23, 9, 18, 36 = ≠1(37) formed by doubling
each conse quetive number, which can be done in your head. 36 is the 18th power of 2
(incidentally this shows that 2 is a primitive root modulo 37), thus we look for the 9th
Normat 2/2013 Ulf Persson 59
power which is 31. Hence 37|1+31
2
which might not have been so easy to find out by
trial and error.
The next step is to do the Euclidean algorithm on n + i and p, which works for
the Gaussian integers.
We have that 13|5
2
+1. Consider
13
5+i
=
13(5≠i)
26
=
5≠i
2
=2+
1≠i
2
thus we can write
13 = 2(5 + i)+(3≠ 2i). Similarly
5+i
3≠2i
=
(5+i)(3+2i)
13
=
13+13i
13
=1+ i hence 5+i =
(1 + i)(3 ≠ 2i) and we get that 13 = (3 ≠ 2i)(3 + 2i)=3
2
+2
2
Note that a
2
+ b
2
is simply the norm of a Gaussian integer, and an example of a
binary form. A similar binary form is given by a
2
≠ab+b
2
which are the norms of the
integers Z[fl] where fl is a primitive cube root of unity. The situation is completely
analogous to the more well-known Gaussian case. A prime is the norm of what I
would prefer to call the Fermat case
0
iff it is of form 6k +1 or 3, in the latter case
it is associated to a square. Whether an arbitrary number can be so represented,
you need to find its prime-decomposition, and ascertain that every prime-factor of
type 6k ≠1 (or 2) occurs with even multiplicity. To find those representations for a
suitable prime p you need to find a non-trivial solution to the congruence equation
x
3
= ≠1 which can be done as before. This means finding a number n such that
p|(n
2
≠n+1) and then to use the Euclidean algorithm to find the greatest common
divisor of p and n + fl in Z[fl] .
37 is a suitable prime in this context as well. Having done most of the work already,
we only need to find the 6th power of two (6=18/3 but 12 would do equally well) which
is 27. We then only need to do the Euclidean algorithm on 27 + fl (or 26 + fl if you prefer)
to end up with the splitting of 37 in Z[fl] . Or just guess and try. 7+3fl works fine as
7
2
≠ 21 + 3
2
=37.
Now because Z[fl] has so many units, in fact six given by ±1, ±fl, ±(1+fl) forming
a regular hexagon in the complex plane, there are many associates to a+bfl. In fact
the following twelve numbers are either associated or associated to the conjugate
of a given a + bfl and thus have the same norm. (The twelve numbers are distinct
except in the case of the norm being equal to 3). Note that a + bfl =(a ≠ b) ≠ bfl
as fl
2
= ≠(1 + fl).
(a,b) (a-b,-b)
(-a,-b) (b-a,b)
(-b,a-b) (-a,b-a)
(b,b-a) (a,a-b)
(a-b,a) (-b,-a)
(b-a,-a) (b,a)
In particular for future reference if we add the two numbers, we get six different
possibilities
±(a + b), ±(2b ≠ a), ±(2a ≠ b)
Now, just as in the Gaussian case a number with many factors will many have
many different representations as a
2
≠ ab + b
2
.
Now if a given number n is the sum of two cubes n = x
3
+y
3
we can factor it into
three factors in Z[fl]. Namely x
3
+y
3
=(x+y)(x+fly)(x+fl
2
y)=(x+y)(x
2
≠xy+y
2
)
0
The ring of integers to the elliptic curve y
2
= x
3
+ ax (associated to the Lemniscate) is the
Gaussian integers, while the ring corresponding to the Fermat cubic y
2
= x
3
+ a is the ring Z[fl].
60 Ulf Persson Normat 2/2013
of which the last two are combined in Z.Thusifn should be the sum of two cubes,
we should be able to factor n = fN where N is a norm, and f = x + y for some
x, y such that Nm(x + fly)=N. Now due to the identity
(x + y)
2
+ 3(x ≠ y)
2
4
= x
2
≠ xy + y
2
we see that f Æ 2
Ô
N,which gives some restriction on the possible N.
We have that 1729 = 7 ◊ 13 ◊ 19. Furthermore 19 = Nm(5 + 3fl), 13 = Nm(4 + fl), 7=
Nm(3 + fl) The possible N are 19 ◊ 7, 13 ◊ 7, 19 ◊ 13. For each of those products we
can find the essentially different representations (modulo those that are generated by the
scheme above). We can represent them in a table.
19 ◊ 7 (5 + 3fl)(3 + fl)=12+11fl (5 + 3fl)(2 ≠ fl)=13+4fl
|a + b| 23 17
|2a ≠ b| 13 22
|2b ≠ a| 10 5
13 ◊ 7 (4 + fl)(3 + fl)=11+6fl (4 + fl)(2 ≠ fl)=9≠ fl
|a + b| 17 8
|2a ≠ b| 16 19
|2b ≠ a| 1 11
19 ◊ 13 (5 + 3fl)(4 + fl)=17+14fl (5 + 3fl)(3 ≠ fl)=18+7fl
|a + b| 31 25
|2a ≠ b| 10 29
|2b ≠ a| 11 4
Due to the two high-lighted values, we see that 1729 can indeed be written as a sum
of two cubes in two different ways.
A similar, if som ewhat more involved, analysis can be made for the case of 4104.
Frequency of sum of two cubes
How common is it that a number is the sum of two cubes? It obviously depends on
the size of the number, the larger, the les s likely. Let us s tart to ask what about
the probability of a number being a square. If the square is about size N,thenext
square is about N +2
Ô
N,thus with a gap of 2
Ô
N.Theprobability of being a
square would then be about 1/2
Ô
N. What about being the sum of two squares? We
consider the numbers N,N ≠1,N≠4,N≠9 ...N≠(
(N/2))
2
and the probability
that each of them is a square. This is paramount to computing the product
1 ≠ (1 ≠ x
1
)(1 ≠ x
2
) ...(1 ≠ x
n
)=
ÿ
i
x
i
≠
ÿ
i<j
x
i
x
j
+
ÿ
i<j<k
x
i
x
j
x
k
...
for some large n. If the first sum is small, we can ignore the other sums. To get an
idea of the first sum we consider the integral as a good approximation of the first
sum.
⁄
Ô
N/2
0
dx
2
Ô
N ≠ x
2
Normat 2/2013 Ulf Persson 61
By making the obvious substitution y =
x
Ô
N
we reduce to an integral we can
actually evaluate (using arcsine).
1
2
⁄
1/
Ô
2
0
dy
1 ≠ y
2
=
fi
8
This is big, which means that we cannot ignore the higher order sums. In fact to
compute the frequency of such numbers is a very delicate matter as shown in a
previous article by Overholdt in a recent issue of Normat. So let us leave this aside
and consider the case of cubes.
A similar argument gives that the probability of being a cube should be
1
3N
2/3
.
This leads to the integral
1
3
⁄
(N/2)
1/3
0
dx
(N ≠ x
3
)
2/3
Making the substitution y =
x
N
1/3
we reduce it to the integral
1
3N
1/3
⁄
1/2
1/3
0
dy
(1 ≠ y
3
)
2/3
This integral cannot be evaluated in closed form, but we can get an approximation
of about 0.88. Now this sum involving the factor
1
N
1/3
is asymptotically so small
that we consider it safe to ignore the higher order terms. The probability of being
the sum of two cubes for numbers the size of N is thus given by CN
≠1/3
where
we have computed the constant C to around 0.3. If we want a formula that gives
the total number of sum of two cubes up to a limit N as say „(N) we should have
for a small number h compared to N that „(N + h) ≠ „(N )=CN
≠1/3
h because
the right hand side counts the numbers in the interval [N,N + h]. Clearly it means
that „
Õ
(N)=CN
≠1/3
, thus that „(N)=C
3
2
N
2/3
≥ 0.44N
2/3
.
This is clearly rather heuristic reasoning. How well does it stand up to reality?
Let us compute this function up to N = 10
6
and compare with our predictions.
The two curves are seen below plotted on a log/log scale. We seen that the rate of
growth has about the right slope and also with the right multiplicative constant.
In particular our formula predicts about 4400 such numbers up to a million, the
actual value is 4454
There is also a more direct and elementary way of estimating the frequency
of sums of cubes. Consider a square with side N
1/3
consisting of numbers (x, y).
62 Ulf Persson Normat 2/2013
Clearly any number less than N involves cubes less than N. On the other side, not
every pair in the square corresponds to a sum less than N , in fact the permissible
pairs need to lie below a certain graph, in fact y Æ (N ≠ x
3
)
1/3
. The number of
such pairs is approximated by the integral
⁄
N
1/3
0
(N ≠ x
3
)
1/3
dx
an obvious change of variables gives us
N
2/3
⁄
1
0
(1 ≠ y
3
)
1/3
dy
Now we only need to approximate this integral and divide by two, as we are counting
each number twice. The result is 0.439N
2/3
which is very close to the one above,
and far easier to motivate and make rigorous.
Now what is the probability that one number should be the sum of cubes in
two different ways? Clearly if the probability of being a sum of two cubes at size
N is given by CN
≠1/3
, being in two different ways should be C
2
N
≠2/3
which
corresponds to a counting function 3C
2
N
1/3
which in our case would correspond
to 0.3N
1/3
which for N = 10
6
would give about 30 coincidences, in reality there
are about 45. Now a million might be too low a number to give a fair reflection of
a trend, but a more extensive count of cases up to a billion shows an even greater
discrepancy. Instead of a predicted 300 of coincidence s one documents 1570 cases.
Could it be that being already the sum of two cubes makes it more likely to be the
sum of two cubes in another way more likely? And if so why?
Now it may come as something of a surprise that one can indeed explicitly solve
the diophantine equation, in the sense of making a complete parametrization of
all solutions. However, this turns out to b e less illuminating and useful, than one
might at first suspect. It all hinges on the geometry of cubic surfaces.
Now the geometry of the cubic surface is classical, and has been the subject
of classical treatieses as well as modern elementary presentations, and usually is
presented as a salivating example in elementary texts on algebraic geometry and
hence there is no need to delve into this in detail in this note, it will suffice to note
the minimum of w hat is needed. For this purpose we start with a short digression
on the geometric construction.
Skew-lines on a Cubic surface
The key-point is to find two skew lines. The existence of lines on a cubic surface
can be shown by abstract formal reasoning (essentially a ’Fubini’type’ argument)
to actually find explicit ones on a given cubic is quite another matter. However,
once one line is found, it is relatively easy to find the others. In our situation, the
form of the cubic will be very simple, and this will not be a problem.
Now if given two skewlines L
1
,L
2
, the lines that intersect them can be para-
metrized by L
1
◊ L
2
which is isomorphic to P
1
◊ P
1
. This is immediate. Given
Normat 2/2013 Ulf Persson 63
any two points P œ L
1
,Q œ L
2
they define a unique line L
PQ
. Now this line will
either be contained in the cubic, or residually intersect it in one point. One can
show that there will be five such lines that are contained in the cubic, provided that
we are willing to look at complex solutions. Conversely given a point outside two
skew lines, there is a unique line through the point intersecting the two skew-lines.
Namely consider the intersection of the two planes formed by the point and either
of the two lines. Or if you prefer, projecting from the given point onto a plane, the
projection of the two skew lines will intersect. Join the point with that intersection
point. Disregarding those exceptional lines lying in the cubic, we will have a 1-1
correspondence between points P,Q (paramatrized by P
1
◊ P
1
and points on the
cubic. Technically one says that there is a birational map between the two varieties.
More explicitly one can easily show that this parametrization can be given by four
cubic forms in three variables, thus defining cubics on the projective plane, passing
through six so called base points.
The special cubic X
3
+ Y
3
= Z
3
+ W
3
Integral solutions to the equation X
3
+ Y
3
= Z
3
+ W
3
are what we are looking
for. Rational solutions are as good, as they lead to integral solutions by clearing
denominators, because if (x, y, z, w) is a solution so is (tx, ty, tz, tw).Thisisjust
an illustration of the fact that the equation is homogenous. Geometrically one con-
siders all of those equivalent, and speaks about the projective space P
3
consisting
of homogeneous four-tuplets (excluding (0, 0, 0, 0)) and with the above equivalence.
Thus among all rational solutions there is one (up to sign) canonical one, which is
a primitive integral solution. That is a solution in integers which have no common
factor.
The form of the cubic is very s imple. It is said to be of Fermat type. It is very
easy to write down all of the twenty-seven lines on it.
Namely considering
(X + Y )(X + flY )(X + fl
2
Y )=(Z + W )(Z + flW )(Z + fl
2
W )
we get immediately 3 ◊ 3=9possibilities
(X + fl
?
Y )=(Z + fl
k
W )=0
using the other two factorizations
(X ≠ Z)(X ≠ flZ)(X ≠ fl
2
Z)=(W ≠ Y )(W ≠flY )(W ≠fl
2
Y )
and
(X ≠ W )(X ≠ flW )(X ≠ fl
2
W )=(Z ≠ Y )(Z ≠ flY )(Z ≠ fl
2
Y )
we get the other 18
Two typical such lines are
X + flY =0 Z + flW =0
X + fl
2
Y =0 Z + fl
2
W =0
64 Ulf Persson Normat 2/2013
where fl is a primitive cube root of one.
Those lines are skew, because if you want to find a solution to the four equations,
you end up with the trivial one (0, 0, 0, 0) which is excluded. Furthermore they are
not defined over the rational numbers Q but over the quadratic extension Q(fl).
Recall that fl satisfies the quadratic equation x
2
+ x +1 = 0,i.e.wehavethe
identity fl
2
+ fl +1 = 0. Now Q(fl) comes with an involution, or if you prefer a
conjugation, given by fl æ fl
2
. This is an automorphism of fields and induced by
complex conjugation. In this sense the two lines are skew. The conjugates of the
points of one line make up the other line.
We will now consider points P on one line and their conjugates
¯
P on the other.
The line L joining the two points will be defined over the reals, as its conjugate
will intersect it in two points and hence coincide. In particular if P œ Q(fl) the line
will be defined over Q. The cubic form above, defined over Q will restrict to a cubic
binary form on L. It will have two conjugate zeroes P,
¯
P , and the third residual
zero (the residual intersection point) will hence be closed under conjugation and
hence belong to Q. Conversely if we have a rational point on our cubic, the unique
line that goes through it, will intersect the two skew lines in conjugate points.
Now we can do this explicitly. The two lines can be parametrized respectively
by (u, ≠fl
2
u, t, ≠fl
2
t) and (v, ≠flv, s, ≠fls),whereu, v, s, t œ Z(fl).
Now we parametrize the line L and it will be given by
⁄(u, ≠fl
2
u, t, ≠fl
2
t)+µ(v, ≠flv, s, ≠fls))
where ⁄, µ œ Z(fl)
Plugging this into our cubic we get the binary cubic in ⁄, µ given by
(⁄u + µv)
3
≠ (⁄fl
2
u + µflv)
3
≠ (⁄t + µs)
3
+(⁄fl
2
t + µfls)
3
Simplifying this we note that (by design) the coefficients for ⁄
3
and µ
3
vanish and
we are left with
⁄µ(⁄(1 ≠ fl
2
)(u
2
v ≠ t
2
s)+µ(1 ≠ fl)(uv
2
≠ ts
2
))
We note that we have two bi-homogenous forms in (u, t; v, s) given respectively
by (u
2
v ≠ t
2
s) and (uv
2
≠ ts
2
) respectively. Bihomogenity is a generalization of
bilinearity, fixing one set of homogenous co-ordinates it will be homogenous in the
other. In fact they will be of bidgree (2, 1) and (1, 2) respectively. (Bilinear forms
have bi-degree (1, 1)). Two such forms will have five intersection points, meaning
that they have common zeroes. For such values the cubic form becomes identicaly
equal to zero, which means that the corresponding lines lie entirely on a cubic.
In fact every two skew-lines on a cubic meet five other lines (necessarily skew).
Two intersection points are obvious, namely (0, 1; 1, 0) and (1, 0; 0, 1) which can be
denoted by (Œ, 0) and (0, Œ) re s pectively. If uv ”=0we can normalize them to 1.
We then see immediately that s = t and s
3
= t
3
=1which gives the three other
lines corresponding to (1, 1), (fl, fl) and (fl
2
,fl
2
) respectively. Thus we see that three
of the lines are real, and the two other complex conjugate. The case of t =1will
correspond to the line L and give a 1-parameter family of trivial solutions given
Normat 2/2013 Ulf Persson 65
by (⁄ + µ, ≠fl(⁄fl + µ),⁄+ µ, ≠fl(⁄fl + µ) from which you c onclude that this is the
lineX = Z, Y = W
Disregarding those five cases, the binary cubic has three solutions, two complex
conjugate and a third, which has to be invariant under conjugation, and hence be
rational.
We get it by setting
⁄ =(1≠fl)(uv
2
≠ ts
2
)
µ =(fl
2
≠ 1)(u
2
v ≠ t
2
s)
So let us denote by A(u, t; v, s) the bihomogenous form (uv
2
≠ts
2
) and by B the one
by switching the variables B(u, t; v, s)=A(v, s; u, t). Furthermore as ⁄, µ are only
defined up to a scalar multiple, it will be convenient to divide them by fl≠fl
2
=
Ô
≠3
so we can write them
⁄ = fl
2
(uv
2
≠ ts
2
)
µ = fl(u
2
v ≠ t
2
s)
We can then get the parametrization of the cubic by
x = fl
2
uA + flvB
y = ≠(fluA + fl
2
vB)
z = fl
2
tA + flsB
w = ≠(fltA + fl
2
sB)
Those will be four forms of bidegree (2, 2) spanning the linear space of all such
form passing through the five intersection points A = B =0. Such a linear subspace
is referred to as a linear system, and the points through which they all pass, the
base points of the system. It is easy to find a basis of 9 forms of bidegree (2, 2)
and each base point imposes one linear condition. They are all independent and
the system will have dimension four, and we have just exhibited an explicit base.
Those four forms give a map from P
1
◊ P
1
to P
4
defined outside the five-base
points, with the image the cubic surface. One says that the five base-points are
blown up to five lines on the cubic.
But now we are interested in the rational solutions, i.e. of Q not just Q(fl).A
sufficient condition for that is that A = B which means that s =
¯
t and v =¯u
We then get
x = ≠|u|
2
+ <u,t>|t|
y = |u|
2
≠ <t,u>|t|
z = |t|
2
≠ <t,u>|u|
w = ≠|t|
2
+ <u,t>|u|
where <u,t>= ≠(fl
2
u
¯
t + fl¯ut)=≠2Re(fl
2
u
¯
t) which is a bilinear form over Q
and satisfies <su,t>=<u,¯st > from which follows that <su,st>= |s| <u,t>
and in particular <u,u>= |u|. We note that <u,t>=<t,u>iff u
¯
t is real or
equivalently t/u is real. In that case we have the trivial solution x + y =0and
z + w =0.
We can now systematically plug in values t/u for elements t, u œ Z[fl] of simple
numerators and denominators, meaning that we bound their norms. We talk about
fractions of bounded heights. By multiplying with a unit in the ring we can assume
66 Ulf Persson Normat 2/2013
that t lies in the sector spanned by 1 and 1+fl. We will only consider examples
when t, u are relatively prime.
We get the following list
15
3
+ 33
3
= 34
3
+2
3
39312 (2)/(≠1 ≠ 2fl)
15
3
+9
3
= 16
3
+2
3
4104 (2)/(1 ≠ fl)
24
3
+ 165
3
= 157
3
+ 86
3
4505949 (2 + 3fl)/(3)
3
3
+ 60
3
= 22
3
+ 59
3
216027 (2 + 3fl)/(3 + 3 fl)
59
3
+ 184
3
= 186
3
+3
3
6434883 (3)/(≠2 ≠ 3fl)
20
3
+ 304
3
= 276
3
+ 192
3
28102464 (3)/(≠1 ≠ 4fl)
20
3
+ 223
3
= 159
3
+ 192
3
11097567 (3)/(1 ≠ 3fl)
86
3
+ 76
3
= 102
3
+ 24
3
1075032 (3)/(1 ≠ 2fl)
90
3
+ 456
3
= 457
3
+ 47
3
95547816 (3 + 4fl)/(2 ≠ 2fl)
108
3
+ 516
3
= 489
3
+ 279
3
138647808 (3 + 4fl)/(4)
168
3
+ 222
3
= 241
3
+ 119
3
15682680 (3 + 4fl)/(4 + 2 fl)
4
3
+ 152
3
= 41
3
+ 151
3
3511872 (3 + 4fl)/(4 + 4 fl)
186
3
+ 1125
3
= 1117
3
+ 332
3
1430262981 (3 + 5fl)/(4 ≠ fl)
352
3
+ 788
3
= 809
3
+ 151
3
532918080 (3 + 5fl)/(4)
420
3
+ 549
3
= 621
3
+ 42
3
239557149 (3 + 5fl)/(4 + fl)
382
3
+ 245
3
= 413
3
+ 16
3
70449093 (3 + 5fl)/(4 + 3 fl)
276
3
+ 180
3
= 297
3
+ 87
3
26856576 (3 + 5fl)/(4 + 4 fl)
87
3
+ 873
3
= 864
3
+ 276
3
665997120 (4)/(≠3 ≠ 5fl)
151
3
+ 617
3
= 620
3
+4
3
238328064 (4)/(≠3 ≠ 4fl)
135
3
+ 825
3
= 760
3
+ 500
3
563976000 (4)/(≠1 ≠ 5fl)
15
3
+ 945
3
= 756
3
+ 744
3
843912000 (4)/(≠5fl)
135
3
+ 633
3
= 508
3
+ 500
3
256096512 (4)/(1 ≠ 4fl)
279
3
+ 297
3
= 360
3
+ 108
3
47915712 (4)/(1 ≠ 3fl)
151
3
+ 425
3
= 332
3
+ 352
3
80208576 (4)/(2 ≠ 3fl)
42
3
+ 504
3
= 378
3
+ 420
3
128098152 (4 + fl)/(2 ≠ 3fl)
119
3
+ 385
3
= 378
3
+ 168
3
58751784 (4 + 2fl)/(1 ≠ 3fl)
47
3
+ 97
3
= 66
3
+ 90
3
1016496 (4 + 2fl)/(4 + fl)
16
3
+ 686
3
= 644
3
+ 382
3
322832952 (4 + 3fl)/(2 ≠ 3fl)
206
3
+ 1180
3
= 1182
3
+ 72
3
1651773816 (4 + 5fl)/(2 ≠ 3fl)
279
3
+ 2241
3
= 2242
3
+ 188
3
11276201160 (4 + 6fl)/(3 ≠ 3fl)
We note that the Ramanujam example does not occur, in fact it will never occur,
because the parametrization only works up to a multiple. Even if the numerator
and denominator of the fraction is reduced, that does not have to be the case with
the image. In fact corresponding to
