CUBIC FORMULA AND CUBIC CURVES

Transcription

1 Commun. Korean Math. Soc. 8 (01), No., pp CUBIC FORMULA AND CUBIC CURVES Sung Sik Woo Abstract. The problem of finding rational or integral points of an elliptic curve basically boils down to solving a cubic equation. We look closely at the cubic formula of Cardano to find a criterion for a cubic polynomial to have a rational or integral roots. Also we show that existence of a rational root of a cubic polynomial implies existence of a solution for certain Diophantine equation. As an application we find some integral solutions of some special type for y = x +b. 1. Introduction We can find a rational solution of an elliptic curve is basically the same as solving a cubic equation. In fact, to find the rational solution of rational cubic equation y = a x +a x +a 1 x+a 0 we need to solve the simultaneous equation { y = a x +a x +a 1 x+a 0, y = αx+β with α,β Q which amounts to solving a cubic equation. In we recall Cardano s cubic formula which gives the zeros of f(x) = x +ax+b. And we show that a cubic f has a rational root if and only if the quantity ω = 7 b+ D, where D is the discriminant of f, is a cube in the splitting field of f. Also we show that existence of rational root of f implies existence of rational solution of a Diophantine equation. In we consideran integralcubic ofthe form f(x) = x +ax+band we find criteria for f to have an integral root in terms of ω. When the class number of Q( D) is not divisible by we give a criterion for f to have an integral root in terms of prime factorization of ω in the ring of integers of Q( D). Received March 10, Mathematics Subject Classification. 11D5, 14G05, 11G05. Key words and phrases. cubic equation, rational solution, integral point of an elliptic curve. 09 c 01 The Korean Mathematical Society

2 10 SUNG SIK WOO In the last section, we consider the zeros of the integral cubic of the form f(x) = x +ax +b. We give criteria for f to have an integral root which are similar to those in. As an application we find solutions of some special type for y = x +b.. Cubic equation The contents of this section are probably well known since ancient times. For completeness we record whatever we need later. To solve a cubic equation y = a x +a x +a 1 x+a 0 we make a change of variable x (X a a ) to get the equation of the form f(x) = X +ax +b Q[X]. Let α,β,γ be the roots of f. The discriminant of f is defined by D = D(f) = (α β) (β γ) (γ α). Also it is well known that the discriminant is given by D = 4a 7b and if D > 0, then f has distinct real roots; if D < 0, then f has one real and two complex roots (conjugate each other). If D = 0, then f has a (real) repeated root and no complex root. Lemma.1. Let α be a root of f(x) be a monic cubic polynomial and let f(x) = (X α)g(x) for some quadratic polynomial g(x). Then D(f) = g(α) D(g). Proof. If β,γ are the roots of g, then g(x) = (X β)(x γ) and D(g) = (β γ). Andg(α) = (α β) (β γ). HenceD(f) = (α β) (β γ) (γ α) = g(α) D(g). Let ρ = 1 + 1, ρ = 1 1 = ρ be two primitive cube roots of unity and 7 A = b+ 7 D, B = b D, where the cube roots are chosen so that AB = p (If b = 0, then A = a,b = a). Then the roots of cubic polynomial f(x) is given by the Cardano s formula [8]: α = 1 (A+B), β = 1 (ρ A+ρB), γ = 1 (ρa+ρ B). We will give the conditions for the cubic polynomials f(x) = X +ax+b Q[X] to have rational roots. We start with an obvious fact:

3 CUBIC FORMULA AND CUBIC CURVES 11 Lemma.. Let f(x) = X +px +q Q[X]. Then f has a rational root if and only if the splitting field of f is an extension of Q of degree. Proof. The cubic f is reducible if and only if f(x) = (X α)g(x) in Q[X] where g is of degree. Hence f is reducible if and only if the splitting field of f is the same as the splitting field of g. And obviously this is equivalent to that the splitting field of g is of degree. Now we want to determine the quadratic extension in the lemma when the rational cubic f(x) = X +ax+b is reducible in Q[X] in terms of the splitting field of f. Proposition.. Let f(x) = X + ax +b Q[X] with D = 4a 7b. Then f has a rational root if and only of the splitting field of f is Q( D). Proof. First suppose D < 0 and f is reducible. In this case, since f(x) is reducible in Q[X] we see that f(x) has one rational root and two complex roots which are conjugate. Let α Q be a rational root of f. Then we can write f(x) = X +ax +b = (X α)g(x) where α Q and g(x) is a monic quadratic rational polynomial with D(g) < 0 by Lemma.1. Also note that irreducibility of g implies b 0 and hence α 0. Let α,β,γ = β be the roots of f. Let β = g + h, γ = g h (g,h Q, h < 0). Then D = (α β)(β γ)(γ α) = h ( h (α g) ). Since α+β +γ = 0 we have α = (β +γ) = g which means that the rational root determines the real part of the two complex roots. On the other hand, since αβγ = b we have α(g h) = b. Hence we have h = αg +b. α Thus D = h[h (α g) ] = h(h 9g ). Now the expression in the square bracket of the last term is a rational number. Hence D (1) h = (h 9g ) = 1 (β γ). Hence α,β,γ Q( D). Next suppose D > 0. Let f(x) = (X α)g(x) with α Q and g(x) is a monic quadratic rational polynomial with D(g) > 0. Let α,β = g + h,γ = g h (g,h Q,h > 0). Then similar computation yields the equality (1) which also shows that α,β,γ Q( D). If D = 0 and if f(x) = X + ax + b Q[X] is reducible, then by direct computation, we can easily see that f has three rational roots.

4 1 SUNG SIK WOO Conversely assume the splitting field of the cubic polynomial f(x) = X + ax +b Q[X] is Q( D). If D is a square in Q, then the splitting field of f is Q which says f has three rational roots. Hence f is reducible of course. If D is not a square in Q, then the Galois group of f is cyclic of order whose generator permutes two roots and fixes the other one. Thus f is reducible in this case too. We saw that the cubic equation f(x) has a rational root if and only if the splitting field of f is Q( D). We will show that this is equivalent to that A,B Q( D). Theorem.4. Let f(x) = X +ax +b be a rational cubic polynomial. Let ω = 7 b+ D with D = 4a 7b. Then f has a rational root if and only if ω is a cube in Q( D), i.e., A Q( D). In this case, if α Q,β,γ are the roos of f, then A = x+y D is given by () x = (β +γ) 1, y = (β +γ)(β +γ). Proof. First consider the case D = 0. In this case, it is easy to show that f is reducible in Q[X] if and only if f has three rational roots. And the latter condition is equivalent to A = B Q. Now consider the case D 0. Assume f Q[X] is reducible and let α,β,γ Q( D) be its roots. First suppose D < 0. Then A,B are real numbers. Suppose α Q and β,γ are complex conjugates. And On the other hand, Hence we have β = ( 1 + i ) A+ ( 1 i = 1 (A+B)+ i(a B). β = (β +γ)+ (β γ). A = (β +γ)+ (β γ), () B = (β +γ) (β γ). Therefore A,B Q( D). Now suppose D > 0 and α,β,γ are three real roots. We see that A,B are complex conjugates say by De Moivre s law. Since f is reducible, we may assume α Q. Similar computation yields the same equation (). ) B

5 CUBIC FORMULA AND CUBIC CURVES 1 Conversely suppose A,B Q( D). And let α,β,γ be three roots of f. Then since 7 b+ D and 7 b D are cubes in Q( D) we can write A = σ +τ D, B = σ τ D with σ,τ Q. Then is a rational root. If D < 0, then α = 1 (A+B) = σ β = ρa+ρ B ( = 1 ) + i (σ +τ ( D)+ 1 ) i (σ τ D) = σ+τ D. Similar computation shows γ = σ τ D. If D > 0, then β = ρa+ρ B = σ τ D and γ = σ + τ D. Hence in either case, we conclude that α Q and β,γ Q( D). Hence we showed that f is reducible if and only if A,B Q( D). But the latter condition is equivalent to A Q( D). For the last part, by (1) and () using the notation of the proof of the proposition above, we get D A = g+ (h 9g ). Using the identity we obtain our result. ( ) ( ) β γ β +γ h =, g = Corollary.5. Let f(x) = X +ax+b Q[X] and let ω = 7 b+ D. If b 0, then f(x) has three rational root if and only if (1) ω = 7 b+ D = (x+y D) for some x,y Q and () D = d for some d Q. If b = 0, then f has three rational roots if and only if () holds. Proof. Clear from the proof of theorem. In [7] we saw that reducibility of a polynomial is equivalent to existence of a common solution of the Diophantine equations. In case of cubics we have:

6 14 SUNG SIK WOO Corollary.6. Let f(x) = X + ax + b Q[X] with D = 4a 7b. Assume D is not a square in Q. Then f = 0 has a rational solution if and only if the Diophantine equations X 9DXY = 7 (4) b, X Y DY = 1 have a common rational solution. Proof. Since f is reducible if and only if ω = 7 b + D is a cube in Q( D), i.e., ω = (x+y D) with x,y Q. Now simply observe that 7 b+ D = (x+y D) = x 9Dxy + D(x y Dy ). If we take the differences we have: Corollary.7. Let D = 4a 7b with a,b Q and D is not a square in Q. Then the cubic X X Y 9DXY +DY = 1 (7b+1) has a rational solution if ω = 7 b + D is a cube in Q( D). The solution is given by () of Theorem.4. Remark. Let D, b be nonzero rational numbers. Let F(X,Y,Z) = X 9DXY + 7b Z, G(X,Y,Z) = X Y DY 1 Z. Then the cubic curves F = 0 and G = 0 are nonsingular curves of genus 1 with the obvious rational points F(0,1,0) = 0,G(1,0,0) = 0. Also the cubic H(X,Y,Z) = X X Y 9DXY +DY + 1 (7b+1)Z is nonsingular unless D = Cubics with integer coefficients In this section we consider the cubic polynomials f(x) = X +ax +b with integer coefficients and let D = 4a 7b be its discriminant. Let ω = 7 b+ D. To see whether f has a rational root we need to see if ω is a cube in Q( D). The rational solution must an integer since it satisfies a monic integral polynomial.

7 CUBIC FORMULA AND CUBIC CURVES 15 For a square free d the ring of integers A = O K of K = Q( d) is given by [] Z+Z d d,(4), A = Z+Z 1+ d d 1(4) and for n Z we define, the square free part d = d(n) of n, by n = b d for some b Z and d is a square free integer. Lemma.1. Let f(x) = X + ax + b Z[X] with the discriminant D = 4a 7b. Let d be the square free part of D and K = Q( d) with the ring of integers O K. Then (1) ω O K and () f = 0 has an integer solution if and only if ω is a cube in O K. Proof. We know ω O K if and only if the norm and the trace of ω are integers []. For ω = 7 b+ D, we have N(ω) = 7 4 (7b D) = ( a) Z and tr(ω) = 7b Z. Hence ω O K. By Theorem.4, f has a rational root if and only if ω is a cube and since f is monic with the integer coefficients the rational root must be an integer. We want to find the conditions for ω to be a cube in ω O K. To fix our notation we briefly summarize factorization of prime ideals of a ring of integers of a number field from Chapter 1 of []. Let K be a finite Galois extension of Q of degree n with the group G. Let A = O K be the ring of integers. Then any ideal of A can be written uniquely as a product of prime ideals of A. If p is a prime ideal of A, then p Z is a prime ideal say (p) = p Z. On the other hand, if (p) Z is a prime ideal, then (p)a is an ideal which has a factorization into prime ideals say (p)a = p e1 1 pe per r and since K/Q is Galois, all e i s are the same and the residue extension degree f = [A/p i : Z/(p)] s are the same (independs of p i ). Hence if we let r to be the number of primes lying above (p), then (p)a = (p 1 p p r ) e, erf = n. The norm of a prime ideal p is defined by N K/Q (p) = p f where f = [A/p : Z/p]. Further, N K/Q (p)o K = σ Gσp = (p 1 p p r ) ef and N K/Q (b) = N K/Q (b) if b = (b) is a principal fractional ideal. If K = Q( d) (d is square free) is a quadratic extension of Q, then e,r,f {1,}. Let O K = A be the ring of integers. Let p Z be a prime. Then (p)a is a product of prime ideals. Since = erf we have the following three cases: p 1 p where p splits if; e = f = 1,r =, pa = p where p ramifies; r = f = 1,e =, p where p remains prime; e = r = 1,f =.

8 16 SUNG SIK WOO We know that p ramifies if and only if p divides the discriminant of K. The criteria which prime ramifies, splits or remains prime is given in ([], 1.1). Finitely many primes ramifies and about the half of the rest split and the other half inert. We will use a generalization of Eisenstein criterion proved in []. Theorem.. Let A be an integral domain with classical ideal theory and let f(x) = X n +a 1 X n 1 + +a n be a polynomial in A[X]. Let a = p e1 1 per r be an ideal of A with r > 0,e j > 0, assume that a divides each coefficient a j of f(x) and that p ei i exactly divides a n. Assume finally that the greatest common divisor of n,e 1,...,e r is 1, i.e., (n,e 1,...,e r ) = 1. Then f(x) is irreducible. We need a special case of this: Corollary.. Let A be a ring of integers of a number field and f(x) = X n a A[X]. If (a) = p e1 1 per r and (n,e 1,e,...,e r ) = 1, then f(x) is irreducible. The following result is a slight generalization of [1](Lemma 5, p. 54). Lemma.4. Let f(x) = X p a K[X] for a field K and a prime p. Then f is reducible if and only if f has a root in K, i.e., f is irreducible if and only if f has no root in K. Proof. First suppose f is reducible and assume p char(k). Let ζ be a primitive p-th root of unity and let α be a root of f. Then the roots of f are α,αζ,αζ,...,αζ p 1. Since f is reducible write f = gh. The constant term of g is of the form d = α i ζ j. Hence d p = α ip = a i. Write 1 = ix+py. Then a = a ix+py = d px a py. Hence a is a p-power. Thus f has a root in K. Now suppose p = char(k). Then since f = 0 we see f(x) = h(x p ) with h(t) = t a which is separable. In this case, every root of f has multiplicity p. Hence f(x) = (X α) p. Since f is reducible, (X α) i divides f; α i K for some i. If ai+bp = 1 (a,b Z), then α = α ai+bp = (α i ) a (α p ) b K. Hence f has a root in K. The converse is obvious. Corollary.5. Let f(x) = X p a K[X] be reducible. Let ζ be a primitive p-th root of unity. Suppose f(α) = 0. If p char(k), there is an i such that ζ i α K. If p = char(k), then α K and has multiplicity p. Proof. In the proof Lemma.4 when char(k) p, the root of f are of the form ζ i α and at least one of them belongs to K. The other case is obvious. Lemma.6. Let A be a Dedekind domain with a quotient field K. Let f(x) = X p a A[X] where p is a prime. Then f is reducible in K[X] if and only if f has a root in A.

9 CUBIC FORMULA AND CUBIC CURVES 17 Proof. Suppose f is reducible in K[X]. By Lemma.4 above, f has a root say α K. Since α p a = 0 we see α is integral over A. Therefore α A since A is integrally closed. Lemma.7. Let K be a number field with the ring of integer A and let h K be the class number of K. Let a A and let (a) = p e1 1 per r with p e i. Then either there is a unit u of K such that ua is a p-th power in A; or p h K. Conversely, if a A is a p-th power and (a) = p e1 1 per r, then p e i. Proof. Let e i = pǫ i and let a = p ǫ1 1 pǫr r. If a (b) for any b A, then a is a nontrivial element in the class group of K of order p. Hence p h K. If a = (b) for some b A, then ua = b p for some unit u A. For the converse, let a = b p and (b) = p ǫ1 1 pǫr r. Then (a) = p e1 1 per r = (b p ) = p pǫ1 1 p pǫr r. By uniqueness of decomposition of an ideal into prime ideals we see that e i = ǫ i p. As required. Lemma.8. Let K/Q be a quadratic extension with the class number h K and the ring of integers A and a A. Let (a) = p e1 1 pe per r with p i Z = (p i ) and let p 1 p p r. Let p be an odd prime with p h K. Suppose N K/Q (a) is a p-th power and ( ) whenever p i = p i+1 (i.e., p i splits) we have p e i and p e i+1. Then there is a unit u A such that ua is a p-th power. Proof. Let N K/Q ((a)) = p e1f1 1 p ef p erfr r. If p j p j+1, then p ejfj j p ei+1fi+1 j+1 is a factor of N K/Q ((a)) and since the norm is a p-power and f i = 1 or we see that p e j,e j+1. If p i = p i+1, then p eifi+ei+1fi+1 i is a factor of N K/Q ((a)). In this case we assumed p e i, p e i+1. In all cases we have p e i. Hence (a) = (p ǫ1 1 pǫ pǫr r )p. If p ǫ1 1 pǫ pǫr r is not principal, then the class group contains an element of order p which contradict to the fact that h K is not divisible by p. Hence (p ǫ1 1 pǫ pǫr r ) is principal, say equal to (α) for some α K. Hence vα p = a for some unit v A by Lemma.7. Hence v 1 a is a p-th power in K, say α p = v 1 a. That is α is a root of X p v 1 a A[X]; α is integralovera. Thereforeα A and ua is a p-th power in A with u = v 1. By Lemma.1, reducibility of f(x) = X + ax + b Z[X] is equivalent to that ω = 7 b + D is a cube in K = Q( D). And by Lemma.7, ω being a cube depends on the prime factorization of ω = p e1 1 per r. If all the exponents e i s are multiples of, then ω is a cube up to a unit of K = Q( D) under the condition h K. It is well known that a A is a unit if and only if N K/Q (a) = ±1 and the group of units U(d) of the quadratic field Q( d) is given by ([] p. 191).

10 18 SUNG SIK WOO {±1,±i} Z/4 (d = 1) {±1,±ρ,±ρ ρ = 1} Z/6 (d = ) U(d) = {±1} Z/ (d,d ) {±ǫ m } Z (d > 0) where ǫ in the last case is the fundamental unit. Let f(x) = X +ax +b Z[X] with D = 4a 7b and ω = 7 b + D as usual. For an integer n we let π(n) be the set of (positive) prime factors of n. Theorem.9. Let f(x) = X +ax +b be an integral cubic polynomial with the discriminant D = 4a 7b. Let d be the square free part of D and let K = Q( d) with the class number h K. Suppose d < or d =. If h K and the primes of π(a) satisfy the condition ( ) of Lemma.8, then f has an integral solution. Moreover, the solution is given by σ where σ +τ d O K is a root of x = 1 ( 7b+ D). Proof. Let ω = 1 ( 7b + D). We know that f has an integral root if and only if ω is a cube in O K by Lemma.1. Now N(ω) = 7 4 (7b D) = ( a) = N(A ). Hence ω = ua with u = ±1 by the conditions on d; ω = A or ω = A. In either case ω is a cube in O K. Now by Theorem.4, f = 0 has a rational root. Since the rational root satisfy a monic integral polynomial it must be an integer. To find the solution we simply note that A = σ +τ d and B = σ τ d and then α = 1 (A+B) is a rational solution. Since N(ω) = ( a) the prime factors of a lies below the prime factors of ω. If they satisfy the condition ( ) of Lemma.8, then the condition h K together with the fact that N(ω) is a cube implies that ω is a cube in O K. Corollary.10. Under the same assumptions on d,h K and π(a) of Theorem.9, (i) when d 1(mod4) there is an A O K with A = ω if and only if 1a d is a square of an odd integer. (ii) when d,(mod4) there is an A O K with A = ω if and only if X + d Y = a has an integral solution. Proof. When d 1(mod4) we have O K = Z+Z 1+ d. Hence any element of O K is of the form A = (k+1)+ d d = a. Hence N(A) = 1 4 (k +1) if and only if (k +1) = 1a d. As required. When d,(mod4) we have O K = Z+Z d. Hence any element of O K is of the form A = r + s d. Thus N(A) = r + d s = a if and only if X + d Y = a has an integral solution. Remark. When d 1(mod4) it is not hard to see if 4a d is a square of an odd integer. But when d,(mod4) it is not so easy to decide whether

11 CUBIC FORMULA AND CUBIC CURVES 19 there is an integral solution for X + d Y = a. In Chapter 19 of [4] there are many cases when such a Diophantine equation has an integral solution. Theorem.11. Let D = 4a 7b for some a,b Z. Let d be the square free part of D and let K = Q( d) with the class number h K. Suppose d < or d =. (1) If there is an A O K with N(A) = a, then (5) X 9DXY X Y +DY = 1 ( 7b 1) has a rational solution. If ω = (σ+τ D), then X = σ,y = τ is a solution. () If the primes of π(a) satisfy the condition ( ) of Lemma.8 and h K, then there is such an A. Proof. As in the proof of Theorem.9, A = σ + τ D O K such that A = ω where ω = 1 ( 7b+ D). As before, 7 b D = (σ +τ D) = σ 9Dστ + D(σ τ Dτ ). Comparing the first and the last expression we see that there is a common solution for X 9DXY = 7 b, X Y DY = 1. Now take the differences of the equations to get our result. The last statement is similar to the last statement of Theorem.9. Corollary.1. Under the same assumptions (1) or () of Theorem.11. (i) when d 1(mod4) the equation (5) has a rational solution if 1a d is a square of an odd integer. (ii) when d,(mod4) the equation (5) has a rational solution if X + d Y = a has an integral solution. Proof. The proof is the same as Corollary.10. Mordell ([4], p.7) gave a family of cubics of similar type without rational solutions. Theorem.1. The integral equation has no rational solutions if ax +bx y +cxy +dy = 1 a d 4(mod9), b 0(mod), c ±1(mod).

12 0 SUNG SIK WOO 4. Integral points of some elliptic curves In this section we consider cubics of the form f(x) = x + ax + b Z[x] with the same method of the previous sections. Using this idea we can consider the integral solution of y = x +b of some special type namely the solution of the simultaneous equation: { y = x +b, Let y = mx. (6) f(x) = x +ax +b. If we make the change of variable x (X a ), then we have (7) Y = X +px +q with p = a, q = a +7b 7 Any cubic polynomial can bring into the form (6) by a change of coordinate with coefficients in a quadratic extension Q. The discriminant of (6) is given by D = 4a b 7b. We let a 7b A = + a 7b D, B = D the three roots of cubic (6) are then given by α = a + 1 (A+B), β = a + 1 (ρ A+ρB), γ = a + 1 (ρa+ρ B). Now let x +ax +b Z[x] and let ω = a 7b + D. Then since N(ω) = 1 4 [(a +7b) +7D] = a 6 Z and tr(ω) = a 7b Z we see that ω O K. Theorem 4.1. Let f(x) = x +ax +b Z[x],D = 4a b 7b and let d be the square free part of D. Let K = Q( d) with the class number h K. Suppose d < or d =. (1) If there is an A O K with N(A) = a, then the cubic equation f = 0 has an integral solution. If ω = (σ + τ d) with σ + τ d O K, then the solution is given by α = σ a. () If h K and the primes of π(a) satisfy the condition ( ) of Lemma.8 and h K, then there is such an A. Proof. TheproofofthisissimilartoTheorem.9withminorchangeN(ω) = a 6 and we omit it. The corresponding statement to Corollary.10 is:

13 CUBIC FORMULA AND CUBIC CURVES 1 Corollary 4.. Under the same assumptions of (1) or () of Theorem 4.1. (i) when d 1(mod4) there is an A O K with A = ω if and only if 4a d is a square of an odd integer. (ii) when d,(mod4) there is an A O K with A = ω if and only if X + d Y = a has an integral solution. Proof. The proof is similar to Corollary.10 with minor change of N(ω) = a 6. Theorem 4.. Let D = 4a 7b for some a,b Z. Let d be the square free part of D and let K = Q( d) with the class number h K. Suppose d < or d =. If there is an A O K with N(A) = a, then (8) X 9DXY X Y +DY = 1 (a +7b+1) has a rational solution. If ω = (σ+τ D), then X = σ,y = τ is a solution. If h K and the primes of π(a) satisfy the condition ( ) of Lemma.8, then there is such an A. Proof. The proof of this is similar to Theorem.11 and we omit it. The corresponding statement to Corollary.1 is: Corollary 4.4. Under the same assumptions (1) or () of Theorem 4.. (i) when d 1(mod4) the equation (8) has a rational solution if 4a d is a square of an odd integer. (ii) when d,(mod4) the equation (8) has a rational solution if X + d Y = a has an integral solution. Proof. We omit the proof. We apply our idea to get a solution of the equation y = x + b of some special type. Rosen ([], ) gave the necessary sufficient condition for the equation to have an integer solution when b < 1 and h K and gave an explicit solution. We know that there are only finitely many solutions for such equation by Siegel Theorem [5]. Example 4.5. To find a solution of y = x +b we consider some special type namely the solution of y is a integral multiple of a solution of x. For this we solve the simultaneous equation { y = x +b (b Z) y = mx (m Z) which boils down to solving the cubic equation (9) x m x +b = 0.

14 SUNG SIK WOO The discriminant and ω of the cubic (9) is D = 4m 6 b 7b and ω = m6 7b + D OK. The norm of ω is N(ω) = m 1. All the information needed in this example is in [] especially Chapter 1. Consider (10) y = x +9. To find the common solution (10) and y = x we need to consider the cubic f(x) = x 4x +9 where we take b = 9,m =. Then we have ( ) square free D = 1, D = 1, d = 1 =, d 1(mod4). part of D Let K = Q( 9). Then δ K = d = 9 since d 1(mod4) and h K = 4 by [6]. Now O K = Z+Z 1+ 9, ω = , N(ω) = 1. Hence () Z is the only prime lie below the prime ideals in the decomposition of ω. Since d 1(mod8) we see () splits in K. We have decomposition ( () =, 1+ )( 9, 1 ) 9. Let p = (, 1+ 9 ), q = (, 1 9 ),σ = If there is A with A = ω, then we must have N(A) = 4. If we let A = 5+ 9, then A = ω (Theorem 4.1). Then (A) = p a q b with a + b = 4 since N(A) = 4. The possibilities are (a,b) = (0,4),(1,),(,1),(4,0). The pair(,)is impossiblefor(a) = (pq) = (4) which is not true. Since h K = 4 we have p 4 = (16,4σ,σ 4 ) is principal and see if p 4 = (A). Now we compute 16 = AĀ;4σ = 8+ 9 = A 7+ 9, σ 4 = = A 8 9. Thus p 4 (A) but as they have the same norm we see p 4 = (A). Thence (ω) = p 1 which is the required condition ( ) of Lemma.8. Incidentally, our computation shows that the class group of Q( 9) is isomorphic to Z/4. Since x 4x + 9 = (x )(x x ) has roots α =,β,γ = 1± 1, f(x) = 0 has the integral solution x = and the equation (10) has an integral solution x =,y = ±6.

15 CUBIC FORMULA AND CUBIC CURVES Now to the Diophantine equation problem: If we let x = 5,y = 1 6, then ( ) 5 9 A = + Hence the equation = (x+y D) = (x 9Dxy )+(x Dy ) D = D = ω. x 9Dxy = 115, x y Dy =. has solution x = 5, y = 1 6. If we subtract these equations we see that the equation x 9Dxy x y +Dy = 56 where D = 1 has the same solution x = 5, y = 1 6. On the other hand, if we take m =, then D = 7 11; D = 8 11;d = 11 1(mod4). But 4a d = 1 which is a prime. Hence there is no integral solution to x 9x 9 = 0. Remark. The solutions of y = x + b (b > 0) with y = mx satisfies b = x m x x. Hence x b which is far smaller than expected by Hall s conjecture [5] x C ǫ b +ǫ. We plan to apply our method with y = x+m instead of y = mx in which the computation will get more complicated whereas we get more general solutions. References [1] N. C. Ankeny and C. A. Rogers, A conjecture of Chowla, Ann. of Math. () 5 (1951), [] H. Flanders, Generalization of a theorem of Ankeny and Rogers, Ann. of Math. () 57 (195), [] K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, nd ed., Springer-Verlag, New York Berlin, [4] L. J. Mordell, Diophantine Equation, Academic Press, [5] J. H. Silverman, The Arithmetic of Elliptic Curves, nd ed., Springer-Verlag, New York Berlin, 009. [6] N. Sloane, Sequences; A01658, in The On-Line Encyclopedia of Integer Sequences. [7] S. S. Woo, Irreducibility of polynomials and Diophantine equations, J. Korean Math. Soc. 47 (010), no. 1, [8] B. L. Van der Waerden, Algebra, 4th ed., Frederick Ungar Publishing Co., New York, 1967.

16 4 SUNG SIK WOO Department of Mathematics College of Natural Science Ewha Womans University Seoul , Korea address: