On the Equation xyz = x + y + z = 1

On the Equation xyz = x + y + z = 1Author(s): Charles SmallSource: The American Mathematical Monthly, Vol. 89, No. 10 (Dec., 1982), pp. 736-749Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2975835 .

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ON THE EQUATION xyz = x + y + z = 1

CHARLES SMALL Department of Mathematics, Queen's University, Kingston, Ontario, Canada

0. Introduction. Beyond providing results about the specific equation xyz = x + y + z = 1, this paper is intended as a palatable introduction to some general techniques in the study of diophantine equations and elliptic curves, particularly over finite fields.

It is known ([C1], see also [Si1] and [Si2]) that the equation of the title has no solutions in the rational field G. In this paper we discuss solutions in the rings Z/m Z of integers modulo m, and in the finite fields Fq of q = pfn elements, p prime, n > 1. (Recall that the number of elements in a finite field is always a prime power, and conversely for any prime power q = pn there is, up to isomorphism, exactly one field with q elements, denoted Fq. For q = pn and q' = p/n' we have Fq c Fq, if and only if p = p' and nln'; p is called the characteristic of Fq. A good reference for these and other basic facts about finite fields is [IR, chapter 7].)

In Z/2Z the equation has the solution x = y = z = 1; however, in Z/3Z and Z/4Z there is no solution, as one sees at once from the fact that, in these rings, xyz = 1 forces each of x, y, z to be + 1. (For the same reason, there are no solutions in the ring Z of integers.) We show in ? 1 that there are solutions in Z/m Z if and only if m is divisible by neither 3 nor 4. The Chinese Remainder Theorem reduces this to the existence of solutions in Z/pnZ for all n > 1 and all primes p + 2, 3. For fixed p + 2, it turns out that existence of solutions in Z/pnZ for all n is equivalent to existence of solutions in Fp = Z/p Z; this follows from a general lifting lemma as discussed in ?1 below. Finally, we establish the existence of solutions in Z/pZ (p + 2,3) directly, by quadratic reciprocity.

In particular, then, the results of ?1 provide solutions in Fp = Z/pZ for all p $ 3, and therefore in all fields of characteristicp + 3. In ?2 we give a precise count of the number of solutions in Fq, for arbitrary q. It turns out, for example, that although there are no solutions in F3, there are twelve in Fg = F3 (i), i2 = - 1, namely the six permutations of (1, i, - i) and the six permutations of (1 + i, 1-i, - 1). To give a further example, we will see that the number of solutions in Fq for q = lln, n = 1, 2, 3, 4, 5 is 3, 105, 1311, 14685 and 161523 respectively.

The equation under consideration can be read over any field, and except for a few fields over which it becomes singular, it defines an elliptic curve. The terminology is explained in ?2; it turns out that our equation is singular only in characteristics 2 and 13. (Viewing the equation as an elliptic curve is basic to the proof in [C1] that there are no rational solutions. A "direct" proof is given in [Si2].) A fundamental theorem on elliptic curves defined over c (see [C2]) states that even when (as in our case) there are no rational solutions, there are always p-adic solutions-that is, solutions modulo pfn for all n > 1 -for all primes p outside a finite set which can be determined from the equation. Our result in ?1 shows that for the elliptic curve given over c by xyz = x + y + z = 1 the only primes which have to be excluded are 2 and 3. The proof here makes no reference to elliptic curves and is completely elementary; the only tools required are quadratic reciprocity and the lifting lemma already referred to. In ?2 we make the connection with elliptic curves explicit, and invoke rationality of the zeta function to count numbers of solutions in Fq. However, we emphasize that the reader does not need to know anything about elliptic curves to follow the argument. On the contrary, it is hoped that the simple treatment here of an explicit example will whet the appetite for deeper study of the general theory, for example via [IR] (especially chapters 10 and 11), [C2], [H'] (especially chapter IV, ?4), [J], [R], or [T].

Charles Small received his Ph.D. under Hyman Bass at Columbia University in 1969 and has been at Queen's University since 1970. He has published more than a dozen papers in algebra and number theory and coauthored two books, one on the Brauer group of a commutative ring and one on homological methods in commutative algebra. He is a violist and the composer of an as-yet-unperformed string quartet, and he avidly collects (and hereby solicits) serendipitous iambic pentameters.

736



ON THE EQUATION xyz = x + y + z = 1 737

1. Solutions in Z/m Z. We will show:

1.1 THEOREM. The equation xyz = x + y + z = 1 has solutions in Z/m Z if and only if m is divisible by neither 3 nor 4.

Write m = 2aopa . a. l where a0 > 0, P1,.. .Pt are distinct odd primes, and the exponents a,,..., at are positive integers. To solve xyz = x + y + z = 1 in Z/m Z is to solve the congruence xyz x + y + z 1 (mod m), and the Chinese Remainder Theorem ensures that this is equivalent to solving the same congruence modulo 2ao and modulo pa, for all i = 1,. .., t. Since we already know there is a solution modulo 2 and none modulo 3 or 4, it is therefore clear that Theorem 1.1 follows from:

1.2 THEOREM. The equation xyz = x + y + z = 1 has solutions in Z/p'Z for all n > 1 and all primes p + 2, 3.

We will prove 1.2 by reducing to the case n = 1 and then using quadratic reciprocity to show existence of solutions in Z/p Z. First, consider the following attempt to solve xyz = x + y + z = 1: eliminate z to get xy(l - x - y) = 1, and arrange this as a quadratic in y:xy2 + (X2 _ x)y + 1 = 0. The discriminant is (X2 _ x)2- 4x; call this A/(x). Now if the ring R we are working in contains an element x such that

(a) A (x) is a square in R, and (b) we can divide by 2x in R,

then the quadratic formula furnishes y and we have a solution to xyz = x + y + z = 1 in R. This proves:

1.3 LEMMA. Let R be a commutative ring in which 2 is invertible,* and define 1: R -> R by A(x) = (X)2- - 4x. If there is an invertible element x E R for which Al(x) is a square, the equation xyz = x + y + z = 1 has a solution in R.

To apply Lemma 1.3 to R = Z/p Z we use quadratic reciprocity. Good references are [IR] chapter 5 and [Se] I ?3. For convenience, we recall the facts we will need.

Fix an odd prime p, and for each integer m let mi be the class of m modulo p, that is, -i is the image of m under the natural mapping Z -- Z/p Z. Then m is a quadratic residue mod p, and we write (m/p) = 1, if -i is a nonzero square in Fp, and m is a quadratic nonresidue mod p, written (m/p) = -1, if -i is not a square in Fp. If -i = 0 (that is, if pIm) we put (m/p) = 0. The mapping (-): Z -- (0, ? 1) so defined is called the Legendre symbol mod p. Clearly (m/p)

p depends only on mi, that is

m P kp m (m?kP) = (!i~for all k

It is also clear that changing m by a square doesn't change (m/p), that is,

(ink2) = ( m ) for all k.

The other basic properties are:

(i)(mr) m r

*Recall that s E R is invertible if st = 1 for some t E R. Thus to require s E R to be invertible is to require that we may divide by s (= multiply by t) in R. For example, 2 is invertible in Z/pnZ and in Fq, q = pn, for all primes p 4k 2 and all n > 1.



738 CHARLES SMALL [December

(ii) (p) = if and onlyifp- l(mod4)

(iii) (p) = 1 if and only if p ?1 (mod 8)

and (iv) For an odd prime q, (p) = (q) unless

both p and q are -3 (mod 4), in which case

The necessary step to apply 1.3 to Fp is:

1.4 LEMMA. Let p > 3 be prime and define A: Fp -Gp as in 1.3. There exists 0 + x E Gp such that A (x) is a square.

Proof. We will show, more precisely, that x = 1 works unless p -7, 11, 19 or -1 (mod 24); that x =-1 works when p 7 or -1 (mod 24); that x = 3 works when p 19 (mod 24); and that for each p -11 (mod 24) either x =-3 or x =-4 will work. (Here "x = 3" means, of course, "x = 3 Fp etc.)

If p 1 (mod 4) then (- l/p) = 1, and since A(1) =-1 - 22 we may take x = 1. This leaves p 7, 11, 19 and -1 (mod 24). If p 7 or -1 (mod 24), thenp -1 (mod 8), so (2/p) = 1. Since A(- 1) = 23, we can take x = -1 in this case. If p 19 (mod 24), thenp 3 (mod 8) and p 1 (mod 3), so that (2/p) = -1 and (3/p) = -(p73) =-1. Hence, since A (3) = 23 - 3 we can take x = 3. Finally, suppose p 11 (mod 24). Then p 3 (mod 8) and p -1 (mod 3), so that (2/p) = -1 and (3/p) = -(p73) = 1. Now observe that A(-3) = 22 . 3 - 13, and A(-4) = 25 * 13. If (13/p) = 1, then we can use x -3; and if (13/p) = 1, we can usex= - 4. This completes the proof of 1.4.

Combining 1.3 and 1.4, we find:

1.5 PROPOSITION. The equation xyz = x + y + z = 1 has solutions in Fp = Z/pZ for allprimes p +# 3.

To prove Theorem 1.2 (hence also Theorem 1.1) we want to lift solutions mod p (guaranteed by 1.5) to solutions mod pf, n > 1. Because of the criterion given by Lemma 1.3, the essential requirement is that we be able to lift squares. Thus we need:

1.6 LEMMA. Let p be an odd prime and m an integer, not divisible by p. If m is a square mod p (that is, if (m/p) = 1), then in fact m is a square mod pn for all n > 1.

Lemma 1.6 is a very special case of a general lifting theorem, so we will not prove it here; see [Se], Cor. 2 of II ?2, or [BS], Theorem 1 of Chapter 1, ?6. (Note, however, with respect to the first of these references, that the statement of Theorem 1, which Cor. 2 depends upon, should have O < j and 0 < 2k in place of 0 < j and 0 < 2k.) For a more elementary treatment see [IR], Prop. 4.2.3. For the algebraically inclined reader we remark that Lemma 1.6 can be rephrased as follows: an integer m, not divisible by the odd prime p, is a square modulo p if and only if it is a square in the ring Zp of p-adic integers. (Similarly, existence of solutions to xyz x + y + z-1 (modulo pfn) for all n > 1 is equivalent to existence of solutions to xyz = x + y + z = 1 in Zp.) Either of the references [Se] or [BS] cited above in connection with the proof of 1.6 will provide further information about Zp and its relation to the Z/p'Z; for algebraically minded readers we remark that Zp can be defined quickly as limZ/pnZ.

We are now in a position to prove Theorem 1.2.

Proof of 1.2: We saw in proving 1.4 that for any primep > 3 we can choose x e (? 1, ? 3, -4>



1982] ON THE EQUATION XyZ = X + y + Z = 1 739

such that (A (x)/p) = 1. Sincep > 3, both x and 2 are invertible in Z/p'Z. Finally, A/(x) is a square in Z/p'Z by 1.6, so we are done by Lemma 1.3 applied to R = Z/p'Z.

A curious by-product of all this is the observation that (X2 _ - 4x is never a square for o + x E Z: if it were, the arguments above would furnish a solution in (a to xyz = x + y + z-1, contrary to [C,].

In the next section we will lift in a different sense, from Fp = Z/pZ to Fq (q = pn, n > 1), and our results will yield a precise count of the number of solutions of xyz = x + y + z = 1 in Fq* It is certainly natural to ask for a similarly quantitative version of Theorem 1.2: how many solutions does xyz = x + y + z = 1 have in Z/p Z? There is a theory, analogous to what we shall use in ?2 for the finite-field case, which bears on this question (see [H] ?7 and [Ig]), but we shall not pursue this here.

2. Solutions in Finite Fields. We saw (1.5) that the equation xyz = x + y + z = 1 has solutions in the prime fields Fp for any prime p + 3, and hence in any finite field of characteristic + 3. (In fact, as we shall see, F3 is the only finite field in which there is no solution.) In this section we show how the rationality of the zeta function of an elliptic curve can be exploited to count how many solutions the equation has in any finite field of characteristic + 2, 13; the result is 2.7 below. The two excluded characteristics are treated towards the end of the section (see 2.9).

For fixed p, let #n denote the number of solutions (x, y) E F2 = F X F to the equation xy(l - x - y) = 1 (or equivalently the number of solutions (x, y, z) E F3 = F x F x F to xyz = x + y + z= 1), where F = Fq is the field with q = pn elements, n > 1.

The first step toward a computation of #n is the observation that the argument given in ?1 to show # I > 0 for p + 3 can be refined to yield a computation of # 1:

2.1 LEMMA. For any odd prime p, the number #I of solutions to xyz = x + y + z = 1 in Gp is

x- I I

P )

where Al(x) = (X2 _ x)2 - 4x and where (A(x)/p) is the Legendre symbol defined in ?1.

Proof. As in ?1, rewrite the equation as a quadratic in y: Xy2 + (X2 -X)y + 1 = 0. The discriminant is A (x). Each nonzero element of Fp is x- for some integer x E S = (l,...,p - 1). From the quadratic formula we see that each x E S with (A (x)/p) = 0 contributes exactly one solution; each x E S with (A (x)/p) = 1 contributes two solutions; and each x E S with (A (x)/p) = - 1 contributes no solution. Hence each x E S contributes (A (x)/p) + 1 solutions, and the total number of solutions is

E ( ))+ 1)=p - I

+ E (A(x

REMARK: It follows from 2.1 that #I can be zero only if (A (x)/p)= -1 for every x= 1, 2,. . , p - 1. The argument in ?1 (via quadratic reciprocity) shows that this never happens for p > 3; indeed, we saw that for each p > 3, some x in (? 1, ? 3, - 4) satisfies (A (x)/p) = 1. An attractive and important problem is the evaluation of sums of the form PI- l (f(x)/p) where f is a polynomial with integer coefficients.

The next step towards a computation of #n is to "homogenize" the polynomial

f(x, y) = xy(l-x-y)-1 = xy-x2y - xy2 1; that is, we consider the homogeneous polynomial

(x,y,z) = xyz-X2y-xy2 -z3

obtained from f by multiplying each term by the appropriate power of the new variable z. Clearly the number #n of zeroes of f in F2 = F x F is the same as the number of zeroes of f in




F3 = F x F X F, if we count the latter properly. Indeed, if we identify proportional solutions to the equation f = 0 (count two solutions (xl, Yl, z,) and (x2, Y2' z2) as the same solution if x2 = /xX, Y2 = /yY, z2 = /z1, for some 0 + Xe F) then #n is exactly the number of solutions (x, y, z) tof = O with z + 0.

More formally, let Nn be the number of equivalence classes of solutions in F3 to f = 0, where the equivalence relation is proportionality, as above, and where the trivial solution (x, y, z) = (0, 0, 0) is not counted. Then, to use the traditional geometric language, Nn is the number of points on the projective curve over F given by the equation f = 0; the points with z = 0 are " at infinity," and those with z + 0 are said to be "at finite distance," or to lie on the affine curve over F given byf= 0.

One may well ask, why worry about the number Nn of points on the projective curve f = 0 when what we are interested in is the number #n f points on the affine curve f = 0. One answer is that some powerful theorems are available for computing Nn, and the number Nn -#n of points at infinity is easily computed. We return to this question at the end of ?3.

How many points at infinity does our curve f = 0, f(x, y) = xy - x2y - xy2 - 1, have? They are the solutions to xyz - x2y - Xy2 - Z3 = 0 with z = 0, that is, the solutions to xy(x + y) = 0 (where "solution" means "proportionality class of solutions, with x = y = z = 0 excluded"). Clearly there are exactly three: x = 0, y = 0, and x =-y; that is, (x, y, z) = (0, 1,0), (1,0,0), and (l,-l,0). Thus:

2.2 LEMMA. #n = Nn- 3. In particular,

I P x~- I (xP) N1~~~X=

where A(x) = (x2 _ x)2 - 4x.

Our next task is to describe the theory used to compute Nn. For this we start with any polynomialf in, say, m variables, with coefficients in Fp. We will define the zeta function Z(t) of f, state the theorem which computes Z(t) in the case when f defines an elliptic curve, and show how this applies to the polynomial f(x, y) = xy(l - x - y) - 1.

Let f be the " homogenization" of f, obtained as above by multiplying each monomial in f by a suitable power of a new variable; thus f is a homogeneous polynomial in m + 1 variables over Fp. Let Nn be the number of solutions in F'+ 1 to f = 0, where F = Fq, q = pn. (Again, "solution" in this context means "proportionality class of solutions, with (xI, x2,..., Xm? 1) = (0, 0, 0) excluded.") Then the zeta function of f is, by definition, the power series

Z( t) = exp E Nn tn

n1 n

It is simplest, and quite sufficient for our purposes, to consider Z(t) as a formal power series in t with rational coefficients. In other words, Z(t) is obtained by formal substitution of x = En lI Nn tn /n in exp(x) = Ei_>x /i!. Since the power series x has zero constant term, the result is a well-defined power series over G.

We are now ready to specialize to the case of elliptic curves. If f is homogeneous of degree 3, in three variables, one saysf is a projective plane cubic; if in additionf is nonsingular, one saysf (orf) defines an elliptic curve over Fp. The condition for nonsingularity of f is that no point (x, y, z) E F3 should satisfy:

(x,y,z) + (0,0,0),J(x, y, z) = (x, y, z) = (x, y, z) =fZ(x,y,z) = O, where F is the algebraic closure of F, and fx, fyJ fz are the partial derivatives of f with respect to x, y, z respectively. (This is the "Jacobian criterion" for nonsingularity; the requirement is simply that at no point on the curve do all the partial derivatives vanish; intuitively, the meaning is that at each point on the curve there is a well-defined tangent direction.)



1982] ON THE EQUATION xyz = x + y + z = 1 741

In our example f (x, y) = xy(1 - x - y) - 1 this is all particularly easy: we have f = xyz - x2y - xy2 z3sothat

Jx = y(z - 2x -y), = x(z - x - 2y),fz = xy - 3Z2. Suppose P = (x, y, z) is a singular point on the curve (that is, a point where f , fx, fy f all vanish). P cannot be one of the points at infinity, for if z = 0 the vanishing of fz shows x = 0 or y = 0, and the remaining conditions give (x, y, z) = (0,0,0). Once we know z + 0 we may assume z = 1, and the conditions fx(x, y, z) = fy(x, y, z) = 0 become

()2x +y = I x + 2y = 1

If p = 3, it's clear that (*) has no solution so f is nonsingular in this case. For any p + 3, (*) has the unique solution x = y = 1/3, and either of the remaining conditions f = 0 or fz = 0 shows that we must have p = 2 or p = 13. Thus:

2.3 PROPOSITION. The equation xy(l - x - y) = 1 defines an elliptic curve over any field of characteristic p + 2,13. When p = 2,(x, y) = (1, 1) is the only singular point, and when p =

13, (x, y) = (9, 9) is the only singular point. (In either case, the singular point is x = y = 1/3.)

The remarkable theorem about the zeta function of an elliptic curve is:

2.4 THEOREM. The zeta function Z(t) of an elliptic curve over Fp, as defined above, is a rational function of t. More precisely, put c = N, - p - 1, then

(I-t)(l-pt) 1-(p + I)t + pt2

For our curve xy(l - x - y) = 1 we have already seen how p determines c:

2.5 LEMMA. LetA(x) (X)2-4x, then c = I + E( ).

Proof. Compare the definition of c in 2.4 with the computation of N1 in 2.2.

For proof and discussion of the fundamental result 2.4 the reader is referred to [J], IX especially ?3, and [R], IV especially ?2; see also [I-R], XI. It should be noted that this result, powerful as it is, is actually just the tip of a large iceberg. We have defined the zeta function Z(t) only for a curve (i.e., a single polynomial f of two variables), but it can be defined for any variety over a finite field. The analogue of 2.4 is true in the general case: the zeta function of any variety is a rational function, of a fairly specific form. Of course this explicit form is more complicated in higher dimensions than in the case of a curve,-indeed, it is more complicated for curves in general than for elliptic curves. However, even in the most general case, as we shall see is true in the elliptic curve case, the zeta function encodes, in an extremely efficient way, all the information about Nn as a complicated but predictable function of n.

The point of Theorem 2.4 is that it allows us to use our knowledge of c (equivalently, NI) to compute Nn for all n. To see this, let B be a (complex) root of the numerator 1 + ct + pt2 of Z(t) and put a = 1/A,s then

1 + ct +pt2=(- at)(l Pt

In terms of a (which is determined by c, hence by p (2.5)), the Nn are given by:

2.6 THEoREM. Forp + 2,13 and for all n > 1, we have Nn = pn + 1 - an-.




Note that -a - (p/a) = c, so that the special case n = 1 of 2.6 is just the definition of c. Putting 2.6 together with 2.2 yields:

2.7 COROLLARY. The number #n of solutions to xyz = x + y + z = 1 in Fq, q = pn, p 7 2, 13, is given by #n = pn - 2 - a - (p/l)n with a determined from p as above:

c= 1 +Pf ( ( I))i+ ct +pt2 = (1-at)(l _ t)

Thus, once 2.6 is proved, we have in 2.7 an explicit computation of #,n. Proof of 2.6. From 2.4 we have

at) I a )Pt n = exp N

hence

EN = log(l - at) + log(l _ P t)- log(l -t) - log(l - pt). n>I

Since

x2 3 log(l + x) = x - 2 + L -

2 3 we have -log(l -x) = Enl>Ix/n. Applying this to x = at, (p/a)t, t, and pt, we have

n>l n n>I n n>l n n na n

which gives the result upon comparing coefficients of like terms. A parenthetical observation: if we differentiate the relation

N n - lg l o g t ) pt 2

on both sides with respect to t, we find that the power series En) Nn tn- is a rational function in t. By an ancient and well-known theorem (see for example [S]) a power series is a rational function if and only if its coefficients eventually satisfy a linear recurrence relation. Thus the Nn, and hence also the #n, satisfy a linear recurrence relation, which can be found explicitly by computing the rational function

d I + ct + pt2 log

dt (1- t)(l-pt P -

We leave the details to the reader. We want next to illustrate Theorem 2.6 and its Corollary 2.7 with a few numerical examples.

Here is a table giving values of c for the first twenty-five primes (with thanks to Wendy Vincent for the computations):

p 1 315 71 11 1317 19 23 29313 7 41

c 4-13 I1 -6 -1 3 -2 0 -6 4 7 0

p 431 47 53159 61 67171 73 791 83 89 971101 c 1 -3 0 1 6 -8 -14 3 -2 -8 -12 6 10 12




To indicate how this information determines the # n for each of these primes, consider first p = 3. We have c = - 1, so to determine a we use

1 - t + 3t2 = (1- at)(l - -t)

Thus a = (1 + -11)/2 will do. Using

N1 = 3 + 1 - a - 3

and

1 = 9 + a(3)

(from 2.6) we find N1 = 3 and N2 = 15. Remembering to discount the three solutions at infinity, we get #l = 0 (as we already knew: xyz = x + y + z = 1 has no solutions in F3) and ?2 = 12 (as indicated in the Introduction).

For p = 11 we have c =-6, and from

1 - 6t + 11t2 = (1 - at)(l - t

we find that a = 3 + i12 will do. Then

a2 = 7 + 6i12, a3 = 9 + 25i12, a4 = -23 + 84i12, a5 = -237 + 229i12,

and using

N = lln + 1 - an -

yields N1 = 6, N2 = 108, N3 = 1314, N4 = 14688, N5 = 161526.

Again remembering to subtract 3 from each of these numbers to allow for the points at infinity, we find the numbers # 1, . . ., #5 indicated in the Introduction. The three solutions in the prime field F 1I are the permutations of (x, y, z) = (7,7,9) = (-4, -4, - 2).

The reader will have noticed that in both of these examples (p = 3, 11) we have p/a = a-, the complex conjugate of a. This is, of course, no accident! Indeed, from the equations

(1-at)(l _Pt) = 1 + ct + pt2 = (1- vt)(l - t)

it is easy to see that p/a = a- a it R (the reals). In fact:

2.8 PROPOSITION. The following are equivalent:

(i) iai= V'i p-

(ii) -a = R (iii) a ? R (iv) c2 4p < 0 (v) 1 + ct + pt2 has no real root. (vi) ici < 2Vp (vii) IN,-p-11<2Vp

The equivalence of (iii) through (vi) results from the fact that c2 -4p is the discriminant of the quadratic polynomial 1 + ct + pt2 of which 1/a is a root, and (vi) X (vii) comes from the definition of c (2.4).




A celebrated theorem due to Hasse (1936) states that the conditions listed in Prop. 2.8, besides being equivalent, all hold true. It then follows from 2.6, since

Nn = pn + I _ an_(P)=pn + 1 afn - ji,

that

IN -pn- lj.jjn + IIn = 2p n/2

Thus to the list of conditions in 2.8 we can add:

(viii) INn - pn - 11 < 2 pn/2 for all n > 1, with strict inequality for n = 1.

The term "Riemann hypothesis" is used in the literature both for statement (i), lal = Vp, and for the equivalent statement (viii) that

IN0 -pn - 11 < 2pn/2.

In the latter form it is a strong estimate for the number of solutions: Nn is within 2p"/2 of pn + 1. In terms of the number #n = Nn- 3 of affine solutions to our equation xy( - x - y) 1,

(viii) can be written as follows:

#n -pPn + 21 < 2pn/2.

One trivial consequence of this is that forp = 3 and n > 1, #n is not 0. With 1.5, this justifies the statement that F3 iS the only finite field in which the equation xyz = x + y + z = 1 has no solution.

Hasse's theorem was generalized in 1947 by Weil to the case of arbitrary curves, and formed the basis for one of the famous "Weil conjectures," settled recently by Deligne. The fact in question,-in our case, the fact that lai = Vp,-is known as the Riemann hypothesis (for varieties defined over finite fields). The classical Riemann hypothesis, of course, is the assertion (as yet unproved) that every complex zero of the Riemann zeta function

~(s) = E, n--S n?> I

satisfies Re(s) = 1/2. To begin to see the connection between these two (one a theorem and one a conjecture, and both referred to as "Riemann hypothesis"), observe that the function 3 of a complex variable s defined by

d(s) = Z(ps) - (pS - a)(Ps -a) (pS5 - l)(p - p)

has all its zeroes on the line Re(s) = 1/2, for if s = a + it and a(s) = 0 we have

VP = lal = IPSI = Pa

so that a = 1/2. It turns out that a(s) can be expressed as

3(s) = EN(d) d>O

where the sum is over all "positive divisors" d on the curve, and for each d, N(d) (the "norm" of d) is a certain power of p; the analogy between 3(s) and the classical Riemann zeta function is then clear.

Forp = 13 the general machinery described above doesn't apply for our curve xy(l - x - y) - 1, since the equation becomes singular in characteristics 2 and 13. However, the value c = -

(computed as in 2.5) yields the correct value N1 = 13; the ten (affine) solutions to xyz = x + y + z = 1 in F13 are (9, 9, 9), three permutations of (2, 2, 10), and six of (1, 5, 8). For n > 1, the formula

Nn = 13n + 1 -_ -_




breaks down; for example it predicts N2 = 195 whereas in fact N2 = 169. Because our curve is singular in characteristic 2 and 13, additional techniques are needed to compute Nn in these two cases. For the sake of completeness we give the results, though without proof:*

2.9 THEOREM. For p = 2 we have N, = 2' if n is even, 2 n+ 2 if n is odd. For p = 13 we have Nn = 13". Thus the number #n of solutions to xyz = x + y + z = 1 in Fq, q = pn, is 2n- 3 when p = 2 and n is even, 2n-l when p = 2 and n is odd, and 13n-3 when p= 13.

To close this section we list a few additional facts, leaving proofs as exercises:

2.10 PROPOSITION. Suppose the elliptic curve E and the prime p are such that c 0. (Since - c = a + a, c = 0 is equivalent to Re(a) = 0; for our curve given by xy(I - x - y) = 1 we have seen that it happens, for p < 101, exactly for p = 23, 41 and 53.) Then Nn = pn + 1 + e where e = 0 if n is odd and e I(-1)m 12pm if n 2m is even.

2.11 PROPOSITION. In general Nn I -(-C)" (mod p). Equivalently, (a + ii)n an + a"

(mod p).

3. Normal Forns. We have, following [R], defined an elliptic curve as an absolutely nonsingular projective plane cubic curve. Elliptic curves have a rich history, and are studied in the literature at various levels of generality; for an indication of how extensive the literature is, see the bibliographies of [C2] and [T]. More general definitions, for example, which we will not pursue here, are that an elliptic curve is a curve of genus 1, or a certain kind of group scheme, or an abelian variety of dimension I (see [H'], IV ?4; also [T]). An alternate (and more down to earth) definition [Sch] which we do want to pursue here briefly, in connection with our curve xyz x + y + z = 1, is that an elliptic curve over a field F is given by an equation of the form y2 = f(X) where f(x) is a polynomial of degree 3 or 4 with coefficients in F and with no repeated roots in an algebraic closure of F. (In this approach, equations y2 =f(X) with f(x) of degree > 4 are "hyperelhptic curves.")

Notice that if we homogenize y2 = f(x) (where f has degree > 3) and then set the new variable equal to 0, we find x = 0, so that any curve of this form has exactly one point at infinity, namely (x, y, z) = (0, 1,0).

We have almost seen in ?2 that our curve xyz = x + y + z = 1 is equivalent (via a suitable change of variables) to one of this type. Indeed, if xy(l - x - y) = 1 and we let

y 2xy + x2 _ x

then

y,2 = (x)= x4-2x3 + x2-4x.

(This can, of course, easily be verified; to see where it comes from, arrange the equation (as before) as a quadratic in y,

xy2 + (X2 - X)y + I = 0, use the quadratic formula to write

y-(_X2?+ x ? A(x)/2x,

and solve for A(x).) Conversely, if Y'2 = A(x) and x * 0, then putting

y = (y' + x - x2)/2x

gives xy(l - x - y) = 1. The two recipes are mutual inverses, and it follows (in all characteristics + 2) that solutions to xy(l - x - y) = 1 are in one-to-one correspondence with solutions other than (0, 0) to y'2 = A(X).

*J thank Alain Robert and Leslie Roberts for some guidance here.




Is this new equation y'2 = A (x) an elliptic curve in Schmidt's sense? To see this we must check whether A(x) can have repeated roots in an algebraic closure of F.; and we can expect that the answer will depend on p. Now the condition for a polynomial to have distinct roots is that it and its derivative have no nontrivial common factor, or, equivalently, that its discriminant be nonzero; and (provided the characteristic is not 2 or 3) the discriminant of a fourth degree polynomial

x4 + blx3 + b2x2 + b3x + b4

is I- 27J2 where

blb3 b22 abb2b4bb2b3 b2 b2b4 b3 4 4 12 6 48 16 16 216

(see [B], II p. 84 and I p. 121). For our polynomial A/(x) we have

b= -2, b2= 1, b3 = -4, b4 = 0,

which gives (incredibly!) I -23/12, J =-181/216, and therefore 3- 27J2 = -26. Hence, except in characteristics 2, 3 and 13, the equation

y2 = A(X) = X4-2x3 + x2 - 4x

defines an elliptic curve in Schmidt's sense as well. Note that the number of points on the projective curve obtained by homogenizing y'2 = A(x),

being one more than the number of points on the affine curve y'2 = A(x), is two more than the number of points on the affine curve xy(1 - x - y) = 1, and is therefore one less than the number of points on the projective curve obtained by homogenizing xy(I - x - y) = 1 (cf. 2.2). Why do these two projective curves, one corresponding to xy(l - x - y) = 1 and the other to y/2 = A(x), have different numbers of points, even though, on the affine level, the simple change of variables

(x', y') = (x,2xy + x2- x)

takes us from one to the other? The answer is that whereas the homogenization xy2 + x2y - xyz + Z3 of xy(l- x - y) -1 is

nonsingular (except in characteristics 2 and 13; see 2.3), the homogenization

1= y2z2 _ x4 + 2x3z - x2z2 + 4xz3

of y2 - X4 + 2x3- + 4x is always singular. Indeed, the partial derivativesfx, fy andfz off are

-4X3 + 6x2z - 2xz2 + 4z3 ,2yz2, and 2y2z + 2x3 - 2x2z + 12xz2,

respectively, and the point (0, 1, 0) at infinity obviously satisfiesf = xfy = fz = 0. (It is the only singular point: at any other point we may take z = 1; the vanishing of fy gives y = 0, and then the simultaneous vanishing of f and fJ shows that A (x) has a repeated root, which we have seen is impossible, at least in characteristics * 2, 3, 13.) In the language of algebraic geometry, the two projective curves in question are birationally equivalent but not isomorphic; the singular point at infinity on one of them corresponds to two distinct points on the other. Although y'2 = A (X) defines an elliptic curve in the sense of [Sch] (in characteristics + 2, 3, 13) the corresponding projective curve, being singular, is not an elliptic curve in the sense of [R]. It is, however, birationally equivalent to the elliptic curve given by xy(l - x - y) = 1. If we want to count the number ?n of points on y'2 = A (X) over Fpn (or the number Mn of points on the corresponding projective curve) we cannot apply the machinery of ?2 directly (but we do have

#n + 1 = Mn = Nn- I

with Nn as in ?2).




Finally, since many discussions of elliptic curves, particularly the classical theory over the field of complex numbers, lay heavy emphasis on the Weierstrass normal form y2 = x3 + ax + b, we want to indicate briefly how our curve looks in this disguise as well.

Given x, y, z such that xyz = x + y + z = 1, put

x' = (z - 12)/3z, y' = 4(x -y)z;

then

t2 t3 23 + 362 Y X 3 27

Conversely, given x', y' satisfying this latter equation, and with x' + 1/3, put x = (3x' - 3y' + 11)/(6x' - 2),y = (3x' + 3y' + 11)/(6x' - 2), andz = 12/(1 - 3x');

then xyz = x + y + z = 1. These two recipes make sense in any field F of characteristic other than 2 or 3, and they are mutual inverses, giving a one-to-one correspondence between solutions (x, y, z) e F3 to xyz = x + y + z = 1 and solutions (x', y') E F2, other than (x', y') =

(1/3, ?4), to

t2 t3 23 + 362 y X 3 x 27

The latter equation defines an elliptic curve in Schmidt's sense provided

h(x) = x3 + 23 x + 362 3 27

has no repeated roots. Again, the criterion is that the discriminant not vanish, and for a cubic polynomial x3 + ax + b the discriminant is 4a3 + 27b2 ([B], II p. 84 and I pp. 72, 83). Putting a = 23/3 and b = 362/27 we find

4a3 + 27b2 = 29 13, so that again 2, 3 and 13 are the only bad characteristics.

Note that the number of points on the projective curve obtained by homogenizing

t2 = h2(x') = x'3 + 33x' + 362 y ~ ~~3 27' being one more than the number of points at finite distance, is three more than the number of points on the affine curve xy(l - x - y) = 1, and hence is equal to the number of points on the projective curve obtained by homogenizing xy(l - x - y) = 1. The projective curve given by the homogenizations of y'2 = h (x') is nonsingular in characteristics + 2, 3, 13: using the Jacobian criterion as above, we find first that the point (0, 1,0 ) at infinity is not a singular point, and then as in the previous case that a singular point at finite distance would provide a common root for h (x') and its derivative, which we know is impossible in characteristics + 2, 3 and 13. In fact the projective curves defined by (the homogenizations of) xy(l - x - y) = 1 and

i2 = 6 23 + 362 Y X 3 27

are isomorphic rather than just birationally equivalent (leaving aside, as always, characteristics 2, 3 and 13); indeed there is a general theorem which states that for nonsingular curves, birational equivalence implies isomorphism. Thus it is no surprise that the two projective curves have the same number of points (in any field of characteristic + 2, 3, 13).

Since y'2 = h(x') is nonsingular (in characteristics + 2,3, 13) the machinery of ?2 can be applied directly to count the number #nt of solutions in Fpn (and the number Nn = # " + 1 of points on the corresponding projective curve). Indeed, the same argument as in 2.1 shows that the number # I' of solutions in the prime field Fp is




x=of P)

and therefore the number # of solutions in Fp to xyz = x + y + z = 1 is fl 1 h (x)

p -2 +E(hG

and the number N1 of points on the corresponding projective curve over Fp is

~~? )

Thus for p + 2,3, 13, the number c = N1 - p - 1 of ?2 can be computed from the Weierstrass normal form as

where

h(x) =x3+ 23 x+ 362 FL~AJA 3 27

The Nn (hence also the #;'= Nn - 1 and the #n = - 2 = Nn - 3 which were the original goal) are then given, in terms of c, by 2.6.

Comparison of these computations for the three curves:

(1) xy(l-x-y) = 1,

(2) y2 = A(X)=X4-2x3 + x2-4x,

and

(3) y2 = h(x) = x3 + 23X + 362 y2==h(x)=x 3 27'

reveals some important general considerations. A simple change of variables, such as the one which took us from (1) to (2), can introduce a singularity and thereby fail to be an isomorphism, and in particular can alter not only the number of affine solutions but also the number of points on the corresponding projective curve. Even a change of variables such as the one which took us from (1) to (3), which preserves nonsingularity and is therefore an isomorphism, can alter the number of affine solutions: by varying the equation we can put more, or fewer, points at infinity, and what is invariant (under isomorphism) is the number of points on the complete (i.e., projective) curve.

This may help explain why the number Nn of points on the projective curve is the more natural concept, and why it is Nn (rather than, say, #J) which appears in the zeta function and plays a role in the general theory. Even if one's ultimate goal is to count numbers of solutions to diophantine equations,-that is, numbers of points on affine curves-over finite fields, it is essential to pass to the projective curve to get a proper picture of what is going on. The reader wishing to go further in this direction is referred to [F] and [0], which provide an introduction to the theory of algebraic curves (and, thereby, an entree into the world of algebraic geometry), as well as the works [C2], [H'], [J], [R] and [T] already cited.

References

[BS] Z. I. Borevich and I. R. Shafarevitch, Number Theory, Academic Press, New York, 1973. [B] W. S. Bumside and A. W. Panton, The Theory of Equations, 7th ed., vol 1: 1912, vol. 2: 1928, Dover,

New York, 1960.



1982] NOTES 749

[C1] J. W. S. Cassels, On a diophantine equation, Acta Arith., 6 (1960) 47-52. [C2] , Diophantine equations with special reference to elliptic curves, J. London Math. Soc., 41 (1966)

193-291. [F] W. Fulton, Algebraic Curves, Benjamin, New York, 1969. [H] S. J. Haris, Number theoretical developments arising from the Siegel formula, Bull Amer. Math. Soc., 2

(1980) 417-433. [H'] R. Hartshorne, Algebraic Geometry, Springer Graduate Texts #52, New York, 1977. [Ig] J. -I. Igusa, Some observations on higher degree characters, Amer. J. Math., 99 (1977) 393-417. [IR] K. Ireland and M. I. Rosen, Elements of Number Theory, Bogden & Quigley, Tarrytown, NY, 1972. [J] J. -R. Joly, Equations et varietes algebriques sur un corps fini, Enseign. Math., 19 (1973) 1-117. [0] G. Orzech and M. Orzech, Plane Algebraic Curves, Marcel Dekker, New York, 1981. [R] A. Robert, Elliptic Curves, Springer Lecture Notes #326, New York, 1973. [Sch] W. M. Schmidt, Equations over Finite Fields an Elementary Approach, Springer Lecture Notes #536,

New York, 1976. [Se] J. -P. Serre, A Course in Arithmetic, Springer Graduate Texts #7, New York, 1973. [Si1] W. Sierpifiski, On some unsolved problems of arithmetics, Scripta Math., 25 (1960) 125-136. [Si2] , Remarques sur le travail de M. J. W. S. Cassels "On a diophantine equation," Acta Arith., 6

(1961) 469-471. [S] C. Small, When is a power series rational?, preprint, 1979. [T] J. T. Tate, The arithmetic of elliptic curves, Invent. Math., 23 (1974) 179-206.

NOTES

EDITED BY SHELDON AXLER, KENNETH R. REBMAN, AND J. ARTHUR SEEBACH, JR.

Material for this department should be sent to Professor J. Arthur Seebach, Jr., Department of Mathematics, St. Olaf College, Northfield, MN 55057

A SPACE WHICH WILL NOT GO WHERE ITS PROPER SUBSPACES WANT IT TO GO

MURRAY BELL Department of Mathematics, University of Manitoba, Winnipeg, Manitoba, Canada R3T 2N2

JOHN GINSBURG Department of Mathematics, University of Winnipeg, Winnipeg, Manitoba, Canada R3B 2E9

In this note we describe an example of a pair of topological spaces X and Y such that every proper subspace of X can be embedded in Y, but X itself cannot be embedded in Y. A familiar example of spaces having these properties is a circle X and a line Y. In this context the two key properties of the circle are that it is homogeneous (for every pair of points p, q of the space, there is a homeomorphism of the space onto itself which maps p to q), and it is not homeomorphic to any proper subspace of itself. Here we will exhibit an ordered space X having the same two properties. Then take Y = X - (p} for any point p in X: X cannot be embedded in Y since X is not homeomorphic to any of its proper subspaces, but every proper subspace of X can be embedded in Y since X - (q) is homeomorphic to Y for every q in X.

We recall that every totally ordered set (T, <) may be regarded as a topological space in a natural way; the open intervals in T are taken as a basis for the topology. Such an ordered space T is connected if and only if its underlying ordering is Dedekind complete and contains no consecutive elements, and T is compact if and only if its underlying ordering is complete. (These facts concerning ordered spaces may be found in 3-0 of [1].)

Let us now construct the space X, which is related to a well-known ordered space called the long line. X is in some sense doubly long. We begin by reviewing the construction of the long line.



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On the Equation xyz = x + y + z = 1