On a problem of Graham By E. ERDŐS and E. SZEMERÉDI (Budapest) GRAHAM stated the following conjecture : Let p be a prime and a 1,..., ap p non-zero re

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On a roblem of Graham By E. ERDŐS and E. SZEMERÉDI (Budaes) GRAHAM saed he following conjecure : Le be a rime and a 1,..., a non-zero residues (mod ). Assume ha if ' a i a i, ei=0 or 1 (no all e i=0) is a mulile of hen is uniquely deermined. The conjecure saes ha in his case here are only wo disinc residues among he a's. We are going o rove his conjecure for all sufficienly large, in fac we will rove a sharer resul. To exend our roof for he small values of would require considerable comuaion, bu no heoreical difficuly. Our roof is surrisingly comlicaed and we are no convinced ha a simler roof is no ossible, bu we could no find one. Firs we rove Theorem 1. Le ilo be sufficienly small, qi < q, >o(r1) : A ={a1,..., ar }, l> y1 1I1o is a se of non-zero residues mod. Assume ha for every he number of indices i saisfying a i - (mod ) is less han il -. Then i a iai - r(mod ) e i = 0 or 1, no all e i = 0 is solvable for every r (mod ). This heorem is erhas of some ineres in iself and easily imlies Grahams conjecure in case each residue ossurs wih a muliliciy <11o. To see his observe ha if n'1'0-< z we can sli our se a1i..., a ino wo disjoin ses which saisfy he requiremens of Theorem 1 and hus e i canno be unique for e i a i - 0 (mod ). Now we rove Theorem 1. Pu i1' 1' 0 =6. Firs we rove he following. Now denoe by F(D) he se of all residues of he form Z e i x i and wih x, ED X+Y={x+y ; HEX, yey}. Lemma. Le B(-A, IB ~ > 2I, (JAI =l >b). Then here is a D(- B so ha F(D)I is greaer han 2 ö 2 IDI.

1 24 P. Erdős and E. Szemerédi To rove he Lemma observe ha we can assume ha here is a B,,- B, B,I ::- J,BJ z ha every residue occurs in Br wih a muliliciy a leas r1 2;12. For if no hen a simle argumen shows ha B conains more ha 'í 2 disinc residues and hen by a heorem of Erdős and Heilbronn Z s i a i - r (mod ) is solvable u i E A for all r [1] which conradics our hyohesis. Henceforh we only consider B,. By he heorem of Dirichle o every beb, here is an ineger,<á2 so ha he residue of b b (mod ) is an absolue value ~8 2. We wan o show ha here is a beb, for which his b b (mod ) is an absolue value > 8~. The number of disinc b's in B, is greaer han (B, has a 4n leas S 4 elemens and a mos q of hem are in he same congruence class). Now here are a mos,2 choices for b hus here are a mos 2 disinc b's for which b - b is in he same residue class, hence here are a mos 2 2 = disinc values 8r 1 n of b for which b - b is no greaer han, bu since here are more han disinc 81 1 4 7 b's in B, here is a beb, for which (1) ó < Ib-bl < 52 as saed. Now we are ready o consruc D. We can assume wihou loss of generaliy ha 1 occurs in B, (and is differen from he b which we jus consruced). Now our se D consiss of b [ a2, b's and g, 1's (by our condiions we have a leas h r1 21 /'2 1's and b's). I easily follows from (1) ha he number of sums S - i d d, ED is a leas l_ I ~ J~ (2) ~ ~ 2 O ~1 7 > lq - 2U2 + (b [ Q j2 L,~ ] L (2) follows from b _L S2 ] and 6=q` 11 which roves he Lemma. Uni D from A and aly he Lemma reeaedly. Thus we obain disjoin ses D i, I ~ i ~ r each of which saisfy he Lemma and heir union has a leas LI A (since by he Lemma if () A -- U D ; 1. 8 >2AI=11) we can selec anoher se D, +,).

On a roblem of Graham 1 2 5 Now denoe by F(D) he se of all residues of he form Z a i di by our Lemma d; E D, (4) ~F(Di)j > IDil 28 1 2 Now clearly (5) F ( ṫ U D ij = F(Dj) + F(D2) +... + F(D,). - ))1 By he Cauchy-Davenor heorem [2] Y F U D i r z min, IF(Di)l ) A = P by (), (4) and (5), which comlees he roof of Theorem 1. Henceforh we can assume ha a leas one residue occurs a leas >7 o imes amongs he a's. Wihou loss of generaliy we can assume ha his residue is 1 and ha 1 occurs _->7 o imes. We have o disinguish several cases. Firs assume ::- o. Several subcases have o be disinguished. Firs assume ha all a 's are ~-, 1<a,<...<a_-::-. Le a,+...+ak_-- be he smalles k wih his roery, k<- is easy o see also a,+...+ak+,.< is obvious hus a,+... +ak+(-a,,-a2-... -ak)1 and a,+... +ak+1+(-a,-... - ak+1)1 give wo reresenaions of 0 wih differen P e l. Thus a leas one of he a's are >-. Clearly one canno have wo incongruen a's in (-,) oherwise Za i is clearly no unique. Le -<a_,<. If a _ -- i mus clearly occur wih muliliciy one (since oherwise ::- o- again gives non-uniqueness for Z Ei). Observe ha in his case a,+...+a,--2(--1)~.- since _---2. Le now k be he smalles ineger saisfying -a,+...+ak-- and now a,_,+(-a_)l and a,+...+ak+(-a,-...-ak)l give wo differen IJ values for e i wha is conradicion. Thus we can assume a _ <. Bu hen a,+a _ 7 < and hus we again ge using -a _, res. -a,-a _, ones wo differen values of 2'e i. This disoses he i=r 9 case > TO. Henceforh assume q,,< - 10. Again we have o disinguish several cases. Firs assume ha here are a mos 1 00 residues amongs he a's greaer han

1 2 6 P. Erdős and E. Szemerédi 100 ' Since here are - a 's no congruen 1 here clearly are a leas P 2 +1 a 's greaer han one bu less han 100 ' Their sum is hus greaer han -. Le a,+......+a, he smalles r for which a,+...+a,~- hen also al+...+ar+ar+l< < -+ 50 < hus al+...+a,+(-a,,-...-a,) 1 and a,+...+a,+a,+,+(- P -a,-...-a,-a,+,)1 again give wo differen values for ZEi. Henceforh we can assume ha here are a leas 100 a.'s greaer han 100 and in fac we can assume ha hey are all less han P2 since as we roved in he revious case a mos one a can be greaer han Le now S, be a se of P 2 ~. 100 a's which are congruen one and S 2 a disjoin se of 200 a's which are also congruen one. Le a be one of he residues in (6) P - Clearly 100 1 2 Thus by Cauchy-Davenor Hence JF(aUS,)J - and JF(A-S,-S2 --a)j 2100 50 ~F(aUS,UA-S,-S2 -a)j (7) 0-1+ Y IAI - ISII - IS2I -1 -P 100 200-1. min {,IF(aUS,)J+JF(A-S,-S 2 -a)j} _ Now we again have o disinguish wo cases. Assume firs E a, ~_.z aa, - 11ó ( - 11oh)- 100 As saed reviously we can assume by he heorem of Erdős and Heilbronn ha he number of disinc a's is less han }á hus we can assume a,- P 1/2 Thus by he heorem of Dirichle here is an s - 21 for which 200

On a roblem of Graham 1 2 7 Ifsay :--200 hen sa,+(-sal)1 and 2sa,-(-2sa,)1 give wo reresenaions of D wih differen Z E i. Thus we can assume sal < 200 bu hen sa, can be relaced by sal ones from Sz and since sa, s his again gives wo disinc value of ~Ej. Thus we can assume, aq_wlo. Thus we have a leas --vió>2 a's disinc from 1 which have no been used in (7). By Erdős-Heilbronn (as used before) a leas one of hese a 's have a high muliliciy and hus here is an sa< Thus 100.. ai< 100 since oherwise we could relace sa of he ones by sa and hus we again ge wo disinc values of Z E;. Now we omi from A all he a's occuring in (6) and we obain a new se A'. Using (6) for A' we again ge reresenaion of 0 (T) (we remark ha we can assume ha a, in (7) and aí in (T) are boh - - 100 hus we do no run ou of ones). Adding he wo reresenaions of D we obain our conradicion. References [1] P. ERDŐS and H. HEiLBRoNN, On he addiion of residue classes mod, Aca Arihnreica 9 (1964), 149-159. [21 H. HALBERSTAm and K. F. ROTH, Sequences, Oxford, 1969. (Received February 14, 1974.)