The Maximum Upp er Density of a Set of Positive Real Numb ers with no solutions to x + y = kz John L. Goldwasser West Virginia University Morgantown, WV 26506 Fan R. K. Chung UniversityofPennsylvania Philadelphia, PA 19104 January 11, 1996 Abstract If k is a p ositive real numb er, wesay that a set S of real numb ers is k -sum-free if there do not exist x; y ; z in S such that x + y = kz .For k 4we nd the maximum upp er densityof a k-sum-free subset of the set of p ositive real numb ers. We also show that if k is an integer greater than 3 then the set of p ositive real numb ers and the set of p ositiveintegers are each the union of three but not two k -sum-free sets. 1. Intro duction Wesay that a set S of real numb ers is sum-free if there do not exist x; y ; z in S such that x + y = z .Ifkis a p ositive real numb er, wesay that a set S is k -sum-free if there do not exist x; y ; z in S such that x + y = kz we assume not all x, y , and z are equal to each other to avoid a triviality when k = 2. Many problem in group theory and numb er theory fo cus on sum-free sets. In work related to Fermat's Last Theorem, Schur [Sc] proved that the p ositive integers cannot b e partitioned into nitely many sum-free sets. Van der Warden [W] proved that the p ositiveintegers cannot b e partitioned into nitely many 2-sum-free sets. If S is a subset of the p ositiveintegers we de ne the upp er density S and lower density U n o jS \f1;2;:::;ngj + S ofS to b e the limit sup erior and limit inferior resp ectively of j n 2 Z .If L n kis a p ositiveinteger let U k and Lk denote the supremum of S and S resp ectively U L over all k -sum-free subsets S of the p ositiveintegers. Let f n; k b e the maximum size of a k -sum-free subset of f1; 2;:::;ng and let Gk denote the limit sup erior over the p ositive f n;k . For any k we clearly have the relationship Lk U k Gk . Since integers of n 1 1 the o dd p ositiveintegers are sum-free, L1 . It is easy to show that G1 ,so 2 2 1 L1 = U 1 = G1 = . Roth [Ro] showed that G2 = 0. His results were strengthened by 2 Szemer edi [Sz], Salem and Sp encer [SS], and Heath-Brown [H]. If k is a p ositiveinteger and S is a k -sum-free subset of the p ositiveintegers with x 2 S 1 and y 2 S \f1;2;:::;kxg, then kx y 62 S ,soLk .Ifkis o dd then, since the o dd 2 1 integers are then k -sum-free, Lk = .Ifk6=2iseven, then the set of all p ositiveintegers 2 n o k2 1 1 1 whose mo d k congruence class is in 1; 2;:::; is k -sum-free. Hence Lk for 2 2 k 2 even k . Chung and Goldwasser [CG1] showed that if n 23 then the set of all o dd p ositive integers less than or equal to n is the unique maximum 3-sum-free subset of f1; 2;:::;ng. 1 Hence L3 = U 3 = G3 = . 2 The ab ove density functions have analogs over k -sum-free subsets of the p ositive real num- b ers where k is any p ositive real numb er. If S is a Leb esgue measurable k -sum-free subset of the p ositive real numb ers, we de ne the upp er density S and lower density S ofS to b e u l n o S \0;x] + the limit sup erior and limit inferior resp ectively of j x 2 R where denotes mea- x sure. Let uk and l k denote the least upp er b ound of S and S resp ectively over all u l measurable k -sum-free subsets S of the p ositive real numb ers. Let g k denote the maximum size of a measurable k -sum-free subset of 0; 1]. Clearly wehave lkukgk for any 1 p ositive real number k. Itisobvious that g 1 = and can b e shown that g 2 = 0. Chung 2 and Goldwasser [CG2] found g k for all k 4 and showed that there is an essentially unique maximum set, the union of three intervals: e ;f ] [ e ;f ] [ e ;f ] 1 1 2 2 3 3 where 2 4 2k 2 f = f = f =1 1.1 1 2 3 4 2 4 2 k 2k 4 k 2k 4 2 and e = f i =1;2;3. i i k In this pap er we will nd uk for k 4. We will generalize a result of Rado [R , R1] by showing that for any p ositiveinteger k greater than 3 the p ositive real numb ers and the p ositiveintegers are each the union of three but not two k -sum-free sets and that the p ositive real numb ers and the p ositiveintegers are each the union of four but not three 3-sum-free sets. 2 2. Maximum upp er densityofa k-sum-free set 2 Lemma 1. Suppose k 4 is a real number, c and w arepositive real numbers with c w , 2 k and S isameasurable k -sum-free subset of the positive real numbers which contains c. Then c 2 2 w + ;w 1 w; S \ 2 k k k with equality if and only if 2 2 S \ w; w = 1 w: k k 2 1 Pro of. Let S b e a set satisfying the hyp otheses and supp ose S \ w; w 6= . If x 2 k k 2 1 S \ w; w then 0 <kxw<w and there is a \forbidden pairing" with resp ect to x of k k [kx w; w]: if z 2 S \ [kx w; w] then kx z 2 [kx w; w] but kx z 62 S . Hence 1 S \ [kx w; w] [w kx w] 2.1 2 n o x 1 2 and if y is the infemum of j x 2 S \ w; w then w k k 1 S \ [ky w w; w] [w ky w w] : 2.2 2 1 2 2 c 2 c If there exists x 2 S \ w; w such that kxw w+ then, letting v = S \ w + ;w , 2 2 k k k k k k by 2.1 1 1 2 v [w kx w] < w 1 w; 2 2 k 2 c 2 c so the conclusion of the Lemma holds if ky w w< w+ .Ifky w w w + , there are 2 2 k k k k three cases to consider. 1 1 1 Case i. Supp ose y . Then ky w w yw and, since S \ w; y w = , k k 1 k 2 c 1 v = \ w + ; min w; ky w w + S \ ky w w; w 2 k k k 1 2 c 1 min w; ky w w w + + [w ky w w] : 2.3 2 k k k 2 1 The right-hand side of 2.3 is clearly a maximum when ky w w = w,so k 2 1 2 c 1 1 k +k4 c v w w+ + w w = w 2 2 k k k 2 k 2k k 2 c k +k4+k 4k 1 2 c w = 1 w 2 2k k k k 2 < 1 w: k 3 1 1 2 Case ii. Supp ose <y + . Then y<ky1 and 2 k 1 k k c 1 2 w + v = S \ ; w + S \ yw; kyw w + S \ ky w w; w 2 k k k c 1 1 2 w w + +ky w w yw + [w ky w w] 2 k k k 2 2 c 2 2 c k 2 1 k 1 y + w + + w = 2 2 2 2 k k 2 k k k k k 2 c k 2 c 2 k +4 = w w < 1 w: k 2k k k k k 1 2 2 Case iii. Supp ose y> + and v = a + b + d + e where a = . Then ky 1 > 2 k k k 2 1 2 c 2 S \ ; w , b = S \ yw; w , d = S \ w; w c , and e = S \ w w + 2 k k k k k 2 c; w . Then b y w and by 2.2. k 2 1 d + e 1 w [w ky w w] k 2 2 k 2 2 k = 1 w y w 1 w b: k 2 k k 2 Hence k 2 b + +e 1 w: 2.4 d 2 k 2 c 1 2 Since c 2 S ,if x2S\ w+ ; w then w<kxc< wc but kx c 62 S . By this 2 k k k k forbidden pairing, 2 ka + d w c w: 2.5 k Thus wehave d k d e e k a+ + b+ + + v 2 2 4 2 2 2 1 2 2 1 2 c = 1 1 w c + 1 w + w 2.6 2 k 2 k 2 k by 2.4, 2.5, and the fact that e c.For equality to hold in the Lemma, wemust b e in Case iii and 2.4, 2.5, and 2.6 all must b e equalities, which completes the pro of.