THE SOLUTION OF 3y2 ± 2" = x3

Transcription

1 PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 69, Number, May 978 Abstract. THE SOLUTION OF y ± " = x The diophantine equation STANLEY RABINO WITZ (») y + "y = x, with y = ± is solved. Let 9 = /, where E Reals. Then fi = {a + b + c9\a, b, c E Z) is the ring of integers of Q(9), is the fundamental unit of Ü and ñ is a unique factorization domain (U.F.D.) (see e.g. []). All English letters (except Z, Q and N) will represent elements of Z and all lower case Greek letters elements of ñ. a\aß and (a, ß)Q are read (respectively) as "a divides ß in ti" and "the greatest common divisor of a and ß in ß". Table I is taken from []. Table I The solution of y + v = x for some values of v. v: Solutions (x, y > - v: Solutions (x, y > : no solutions : <, > 4: no solutions 4: < -, 4>, <, 5>, <, >, <858, 76844) 7:<,> 7:<-,> 54: <7, 7> 54: <, 9> 8: no solutions 8: < -, 9>, < -, ), <6, 8>, <66, 7> 6: <6, >, <, 8>, <, 89> 6: < - 6, > 4 : <, 6> 4 : no solutions Note that if (*) holds, then n >, since " = y(x y). Proposition. // (*) holds with n = k and xy odd, then <y,, x, y ) = <,,,>. Proof. Since n = k, (9y) + 7 *y = (x). Using Table I we may assume that k >. By (*), y = ab, where a = x - ky and b = x + *yx + k. Since x is odd, b is odd. b > [, Lemma ] and therefore a >. Received by the editors October 8, 976. AMS (MOS) subject classifications (97). Primary B. Key words and phrases. Ring of integers, norm. 'This material forms a part of the author's doctoral dissertation ("On Mordell's equation y + k = x, with 7c = ±"""') completed in 97 at the City University of New York/Graduate Center. American Mathematical Society 978

2 4 STANLEY RABINOWTTZ a = x -ky Hence x = *y (Mod ), which implies that = x - A:y = y = (Mod ). b = x + *yx + * = * + k + k = (Mod ). Thus 9 a? = y, implying that \y. Also (a, b)\ k [, Lemma ]. Since b is odd, (a, b) =. Since (y/) = (a/)(z?/), [, Lemma ] proves that <a/,?/> = (r, s) or <r, 5>. If <a/, Z?/> = <r, s>, then x = r + *y and we obtain (i - V - Ia) = k, which cannot hold. Therefore <a/, Z?/> = <r, s>. Hence s is odd, x = 9r + *y, and 7r4 + 9 *yr + (* - s) =. 8j - 7 * = d [, Lemma 4]. Consequently a", since k >. Thus d = 8Z> and s - D = k~. Therefore D is odd and it follows that 5 - D = -(Mod 8). But 4 *". Proposition.f(*) holds with n = k + and xy odd, then <y,«,x, y > = <-l,,, >,<-,7, -5, > or <-,, 95, 488). Proof. If k =, then (9y) + 54y = (x). We may assume that k > by using Table I. (x, ) =, for otherwise ". Thus by [, Lemma 5], y = x _ *+iy = ± i ± (Mod 9). Hence (y, ) =. By (*), y = aß, where a = x - kyb and ß = x + ky9x + (k9). ß > [, Lemma ] and thus a >. (a, ß)a\a(k9) [, Lemma ]. Since (aß, )ß = [, Lemma 6], (a, ß)a\a = ( + 9)(- + 9). By [4, (), p. ] and [, Theorem -, p. 48] the norm N of fi (over Z) is A(a + b9 + c9) = a + b + 4c - 6abc. Now A(l + 9) =. Thus + 9 is a prime of ß [, Theorem -, -5, pp. 49, 5]. Also x + *y = x + *y = x - *+y = y = (Mod ). Let w = (x + *y)/. Then a = ( + )(w - ky - w9 + w9). Therefore + 9\aa. If ( + f?) ßa, then 9 = A((l + )) A(a) = y, contradicting ( v, ) =. Since Q is a U.F.D. and is a unit of, (a/(\ + 9), ß)u =. Also ((I + 9)yf= (a/ (\ + 9))(ß/ (-\ + 9)). a/(\ + 9) = p(a + b9 + c9), where p = or [, Lemma ]. We may assume that a - b >, since (a + b9 + c9) = (- a - b9 - c9). If jn =, then since, 9 and 9 are linearly independent over Q, x = a + 4bc + b + 4ac and -*y = a + 4bc + c + az?.

3 THE SOLUTION OF y ± " = X 5 The first equation implies that a is odd and the second that a is even (since k > ). Thus p. = I + and we have the following: () x = -a - 4bc + 4c + 4ab, () -*~'y = -c - ab + b + ac, and () = a + 4»c - t - ac. By (), a is odd. From () it follows that b is odd and (4) c(a - b) = a - b. Multiplying () by 4(a - b) and using (4) we have (5) -*+y(a - A) = (a - b)(a - ab + ab - 5b). Let g = (a, b)>,a = a/g and B - b/g. Thus gb is odd, A - B > and (A, B) =\.(A - B, A - B)=\, since (A - B) - (A - B) = A, and (A - B)-(A - B) = B. (5) yields (6) -k+xy(a - B)= g(a - B)(A - AB + AB - 5B). Hence g (^ - fi). Thus g\a - B. Therefore (A - B, ) = and A + B = A - B = (Mod ). Consequently (A - B, A + B) =, since (A - B) + (A + B) = A and (A + B) - (A - B) = B. It follows from (4) that c (A - B)/g = (A - B)-(A + B)/. Hence A - B = ±g. By (6), (7) - * + y = (A - B)(A - AB + AB - 5B). Therefore A B = r, where < r < k +. (7) implies that - *+-'y = (A - B)- 4B = r - 4B. Thus /-^O. Therefore r >. Hence k + - r = [, Lemma 7] and we have (8) r" + y = B. If r = t, then ('") + y = B. There are no solutions in this case by Table I. Thus r = t + and by (8), (,+f+4y = (B). Using Table I, y = - and (B,,+> = < -, 4> or <, >. If <B,,+> = < -, 4>, then B-, t =, r = t + I = \,A = B + r =, k = + r = and n = k + = 7. Since ±g = A - B = and g >, g =, a = g/4 = and b = gß = -. By (4), () and (*), c =, x = -5 and y =. If (B,,+> = <, ) we obtain similarly, k - 4, x - 95 and y = 488. Proposition. // (*) fows w/ia «= k + and xy odd, then (y,n,x, y > = <-l,, -, > or <-l, 8,, >.

4 6 STANLEY RABINO WITZ Proof. If k =, then (9y) + 8y = (x). The solution of this equation is given in Table I. Thus we may assume that k >. (x, ) =. By [, Lemma 5], v = x - *+y = ± ± 4 (Mod 9). Thus (y, ) =. By (*), y = aß, where a = x-ky9 and ß = x + ky9x + (k9 f. As in Proposition, a > and (a, ß)a\a = ( + 9)(- + 9). Also x - *y = x - *y s x - *+y = y = (Mod ). Let u = (x - *y)/. Then a = ( + 9)(x - u + (m - x)9 + u9). Following the argument of Proposition we see that since N(a) = y, a/(\ + 9) = p(a + b9 + c9), where p = or - + 9, and c >. If p =, then x = a + 4bc +? + 4ac and = a + 4bc + c + ab. The first equation implies that a is odd and the second that a is even. Hence p = + 9. Consequently (9) x = - a - 4bc + 4c + 4ab, () = - c - ab + b + ac, and () -ky = a + 4bc - b - ac. a is odd by (9). Therefore since k >, b is odd by (). () implies () a(b - c) = b- c. Multiplying () by -(b - c) and applying () yields () *y(? - c)= c(c - 6?c + 6?c - Z?). Let g = (b, c) >, B = b/g and C = c/g. Therefore gb is odd, C >, (B, C) = and by (), (4) ky(b -C)= gc(c - 6BC + 6BC - B). As in Proposition, (B - C, B - C) =, g\b - C, and (B - C, B + C) =. Thus (C, ) =. It follows from () that a (B - C)/g = (B - C)(B+ Therefore B - C = ±g. By (4), C)/. *y = C(C - 6BC + 6BC - B). Hence C = r, where < r < k. Thus (5) *"ry = r - r+fi + r+5-5. r =, since k >. Therefore r > r + I > r + I >. By (5), k - r = and (6) r~x +y = (r - B). If r = t, then ('+) + 4y = ((r - B)). There are no solutions of this equation by Table I. Thus r = t + and (6) becomes

5 n _ v THE SOLUTION OF y ± " = X 7 ('+x) + y = (r - B). Using Table I we find that y = -,,+ =, and r - B =. Therefore / =, r = It + =, B = r - =, C = ' =, /c = r + =, and ±g = B - C = -. Hence g =, b = gb =, c = gc =, and by (), a =. By (9), x = and finally y = x - k+y = (). Theorem. /<// iac solutions of (*) are g/uew in Table II, vvaere x = ge and y = ±hf. Explanation of Table II. is given modulo 6 and is nonnegative. If / =, the value of A is irrelevant. The solutions are numbered for reference in the proof y n (modulo 6) 4 (n>l) (n > ) ( > 8) g Table II A - n n «- -8 n n f Solution number Proof. By direct calculation the above can be shown to be solutions of (*). Suppose now that (*) holds. Obviously x ^. If y =, then x /ly and therefore t7. Thus solution,, 5 or holds. Assume that y J=. Therefore x = % and y = hf, where efis odd. By (*), (7) A/ + "y = ge. It follows from [, Lemma 7] that (8) A = g < n, (9) A = < g, or () g = < A. If (8), then A = g = 6w. By (7), / + "-6my = e. Propositions, and imply that solution, 6, 7, 8, 9 or must hold. If (9), then n = 6w + i, where i =, or. By (7),(9 J/) + 7 iy = ( g~we). By Table I, solution 4, or holds.

6 8 STANLEY RABINOWITZ If (), then n = 6w + /', wherey = or. By (7), (9 h-wff+ 7 jv = ( Jef. There are no solutions by Table I. Note added in proof. I wish to thank Professor N. M. Stephens for informing me that the results of my doctoral dissertation, from which this paper and [] are taken, were also determined by Professor F. B. Coghlan in his doctoral dissertation "Elliptic curves with conductor N = mn" completed in 967 at Manchester. References. O. Hemer, On the Diophantine equation y k = x, Thesis, Univ. of Uppsala, Almqvist & Wiksells, Uppsala, 95. MR 4, 54.. W. J. LeVeque, Topics in number theory, Vol. II, Addison-Wesley, Reading, Mass., 96.. S. Rabinowitz, ie solution ofy ± " = x, Proc. Amer. Math. Soc. 6 (9), B. L. van der Waerden, Modern algebra, Vol. I, Ungar, New York, 949. Department of Mathematics and Computer Science, Kjngsborough Community College/ CUNY, Brooklyn, New York 5