Infinitely Many Insolvable Diophantine Equations

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1 ACKNOWLEDGMENT. After this aer was submitted, the author received messages from G. D. Anderson and M. Vuorinen that concerned [10] and informed him about references [1] [7]. He is leased to thank them for this imortant information. REFERENCES 1. G. D. Anderson, S.-L. Qiu, M. K. Vamanamurthy, and M. Vuorinen, Generalized ellitic integrals and modular equations, Pacific J. Math. 19 (000) G. D. Anderson, M. K. Vamanamurthy, and M. Vuorinen, Inequalities for quasiconformal maings in sace, Pacific J. Math. 160 (1993) , Conformal Invariants, Inequalities, and Quasiconformal Mas, Wiley, New York, , Toics in secial functions, in Paers on Analysis, Re. Univ. Jyväskylä De. Math. Stat., no. 83, 001, Jyväskylä, Finland, I. Chavel, Riemannian Geometry A Modern Introduction, Cambridge University Press, Cambridge, J. Cheeger, M. Gromov, and M. Taylor, Finite roagation seed, kernel estimates for functions of the Lalace oerator, and the geometry of comlete Riemannian manifolds, J. Diff. Geom. 17 (198) M. Gromov, Isoerimetric inequalities in Riemannian manifolds, in Asymtotic Theory of Finite Dimensional Saces, Lecture Notes in Math., no. 100, Sringer-Verlag, Berlin, 1986, B.-N. Guo, B.-M. Qiao, F. Qi, and W. Li, On new roofs of Wilker s inequalities involving trigonometric functions, Math. Ineq. Al. 6 (003) I. Pinelis, Extremal robabilistic roblems and Hotelling s T test under symmetry condition (rerint, 1991); a shorter version of the rerint aeared in Ann. Statist. (1994) , L Hosital tye results for monotonicity, with alications, J. Inequal. Pure Al. Math. 3 (00), article 5, 5. (electronic). 11., L Hosital tye rules for oscillation, with alications, J. Inequal. Pure Al. Math. (001), article 33, 4. (electronic). 1., Monotonicity roerties of the relative error of a Padé aroximation for Mills ratio, J. Inequal. Pure Al. Math. 3 (00), article 0, 8. (electronic). 13., L Hosital tye rules for monotonicity: Alications to robability inequalities for sums of bounded random variables, J. Inequal. Pure Al. Math. 3 (00), article 7, 9. (electronic). 14. J. S. Sumner, A. A. Jagers, M. Vowe, and J. Anglesio, Inequalities involving trigonometric functions, this MONTHLY 98 (1991) J. B. Wilker, Problem E3306, this MONTHLY 96 (1989) 55. Deartment of Mathematical Sciences, Michigan Technological University, Houghton, MI iinelis@mtu.edu Infinitely Many Insolvable Diohantine Equations Noriaki Kimura and Kenneth S. Williams Let f (x 1,...,x n ) be a quadratic form in n variables x 1,...,x n with integral coefficients, let be a rime, and let k be a ositive integer. The congruence f (x 1,...,x n ) 0(mod k )issaidtobesolvable nontrivially if there exist integers x 1,...,x n such that f (x 1,...,x n ) 0(mod k ) with at least one of x 1,...,x n not divisible by. Thus the congruence x 1 + x 0(mod3k ) is solvable (with x 1 = x = 0) but is not solvable nontrivially as any solution x 1, x satisfies x 1 x 0(mod3).Letm be a ositive integer larger than 1. The congruence f (x 1,...,x n ) 0(modm) issaidtobesolvable nontrivially if f (x 1,...,x n ) 0(mod k ) is solvable nontrivially for each rime December 004] NOTES 909

2 divisor of m and each ositive integer k such that k is the largest ower of dividing m (written k m). We note that the comonents x i of the solution that are not divisible by are not necessarily the same for different rime divisors of m. The Hasse-Minkowski theorem [1,. 61] asserts that, if (a) there exist real numbers r 1,...,r n not all zero such that f (r 1,...,r n ) = 0 and (b) the congruence f (x 1,...,x n ) 0(modm) is solvable nontrivially for every ositive integer m greater than 1, then the equation f (x 1,...,x n ) = 0 is solvable in integers x 1,...,x n not all zero. However, if f (x 1,...,x n ) is a quadratic olynomial that is not a quadratic form (i.e., is not homogeneous), then (a) and (b) do not ensure that f (x 1,...,x n ) = 0 is solvable in integers x 1,...,x n.anexamleisgivenin[5,. 195]. We give infinitely many quadratic olynomials f in two variables such that (a) and (b) hold but f (x 1, x ) = 0 is not solvable in integers x 1 and x. We make use of a number of elementary arithmetic facts. In (i) (vi) to follow, is an odd rime, a, b, andc are integers, α is a ositive integer, and ( ) is the Legendre symbol defined by ( ) a = 1 if a and x a (mod ) is solvable, 1 if a and x a (mod ) is not solvable, 0 if a. (i) If a 1 (mod 8), then the congruence x a (mod α ) is solvable [,. 13]. (ii) If ( a ) = 1, then the congruence x a (mod α ) is solvable [,. 13], [4,. 137]. (iii) If a, then the number of solutions of the congruence 0 (mod ) is ( ) b 4ac 1 + [3, ]. (iv) If a and b 4ac, then 1 ( ) ( ) a = x=0 [3,. 8]. (v) If D is a ositive integer that is not a erfect square and E is an integer such that 0 < E < D, then the equation x Dy = E is solvable in corime ositive integers x and y if and only if E = h n Dk n for some convergent h n /k n of the continued fraction exansion of D with n in {0, 1,,..., l 1}, wherel is the length of the eriod of the continued fraction exansion of D [4,. 35]. If a and b 4ac we infer from (iii) and (iv) that the number of x in {0, 1,,..., 1} such that ( ) = c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 111

3 is 1 1 x=0 ax +bx+c ( 1 + ( )) = 1 ( 1 ( b 4ac ) ( )) a. When 5wehave ( ( 1 b 4ac 1 ) ( )) a 1 ( 3) 1. Thus we conclude: (vi) If 5, a, and b 4ac, then there exists an integer x such that ( ) = 1. Theorem. If a is an integer greater than 1 each of whose rime divisors is congruent to either 1 or 3 modulo 8, then the equation is not solvable in integers x and y, but the congruence x (a 4 + a )y + 1 = 0 (1) x (a 4 + a )y (mod m) () is solvable nontrivially for every ositive integer m greater than 1. Proof. As (a ) < 4a 4 + a <(a + 1), the ositive integer 4a 4 + a is not a erfect square. The continued fraction exansion of 4a 4 + a is of eriod two and is given by 4a4 + a =[a,, 4a ]. The convergents [4,. 33] of this continued fraction exansion are Let so that h 0 = a k 0 1, h 1 = 4a + 1,... k 1 g n = h n (4a4 + a )k n (n = 0, 1,,...), g 0 = a, g 1 = 1,... December 004] NOTES 911

4 Since < 4a 4 + a, g 0 =, and g 1 =, statement (v) imlies that the equation x 1 (4a4 + a )y 1 = is not solvable in integers x 1 and y 1. Thus the equation (1) is not solvable in integers x and y. Now let m be a ositive integer larger than 1. We show that the congruence () is solvable nontrivially. Let be a rime with m, say α m, whereα is a ositive integer. We consider three cases according as (A) =, (B) = and a 4 + a, or (C) = and a 4 + a. Case A: =. As a is odd, a 1(mod8),soa 4 + a 1(mod8).Thus,by (i), the congruence z a 4 + a (mod α ) is solvable. Then the congruence x (a4 + a )y + 1 0(modα ) is solvable nontrivially with x = y = t,wheret is the inverse of z modulo α. Case B: =, a 4 + a. As a 4 + a we have either a or a + 1. In the former case 1 or 3 (mod 8) by assumtion, so ( ) = 1. In the latter case (a) (mod), so ( ) = 1. According to (ii) there exists in each case an integer w such that w (mod α ). Then the congruence x (a4 + a )y (mod α ) is solvable nontrivially with x = lw and y = 0, where l is the inverse of modulo α. Case C: =, a 4 + a. Because 3 a 4 + a,wehave = 3, imlying that 5. Set b = a 4 + a,sothat b. By (vi) there exists an integer y such that ( ) by = 1. In view of (ii) there exists an integer z such that z by (mod α ). Then the congruence x (a4 + a )y + 1 0(modα ) is solvable nontrivially with x = lz and y = y,wherel is the inverse of modulo α. Finally, aealing to the Chinese remainder theorem, we choose integers x and y such that x x (mod α ), y y (mod α ) for every rime divisor of m and each ositive integer α such that α m. Then x (a 4 + a )y + 1 0(modm). Let f m (x 1, x ) = x 1 ( 34m + 3 m )x + 1 (m = 1,,...). Clearly each equation f m (x 1, x ) = 0 has a nontrivial real solution (x 1, x ) = ( 0,(1/ 3 4m + 3 m ) ), 91 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 111

5 so condition (a) of the Hasse-Minkowski theorem is satisfied. By the theorem condition (b) is also satisfied. On the other hand, none of these olynomials has a solution (x 1, x ) in Z. REFERENCES 1. Z. I. Borevich and I. R. Shafarevich, Number Theory, Academic Press, New York and London, L. E. Dickson, Introduction to the Theory of Numbers, Dover, New York, E. Grosswald, Toics from the Theory of Numbers, Macmillan, New York, I. Niven, H. S. Zuckerman, and H. L. Montgomery, An Introduction to the Theory of Numbers, 5th ed., John Wiley and Sons, New York, W. Sieriński, Elementary Theory of Numbers, Monografie Matematyczne, Tom 4, Państwowe Wydawnictwo Naukowe, Warsaw, College of Industrial Technology, Nihon University, Chiba 75, Jaan nrkmr@math.cit.nihon-u.ac.j School of Mathematics and Statistics, Carleton University, Ottawa, Ontario, Canada K1S 5B6 williams@math.carleton.ca Forward Shifts and Backward Shifts in a Rearrangement of a Conditionally Convergent Series Jón R. Stefánsson In [,. 57] it is roved that a rearrangement of a conditionally convergent series remains convergent (with unaltered sum), rovided the series is rearranged in such a way that the forward shifts are bounded. It is remarked that there is a clear difference between forward shifts and backward shifts. The urose of this note is to show that in the context under consideration there is, in fact, no difference. The result stated in [] holds as well with the assumtion that the backward shifts are bounded. Let x n be a series, and let x π(n) be a rearrangement determined by a ermutation π of the natural numbers. The nth term, x n = x π(π 1 (n)), of the original series is shifted to the kth term of the rearranged series, where k = π 1 (n). The forward (resectively, backward) shifts are bounded if and only if the differences π 1 (n) n (resectively, n π 1 (n)) are bounded above. We state the following theorem, where we do not a riori assume convergence of the series: Theorem. Let x n be a series in a normed linear sace with lim x n = 0, and let xπ(n) be a rearrangement of the series. Assume that either the forward shifts are bounded or the backward shifts are bounded. If s n (resectively, t n ) denotes the nth artial sum of the series x n (resectively, x π(n) ), then the following statements hold: December 004] NOTES 913