SCHUR S THEOREM REU SUMMER 2005 1. Combinatorial aroach Prhas th first rsult in th subjct blongs to I. Schur and dats back to 1916. On of his motivation was to study th local vrsion of th famous quation of Frmat: x n + y n = z n. If thr ar intgrs x, y, z satisfying th abov quation, thn for vry rim, thy also solv th congrunc quation: x n + y n z n (mod ). H showd that th congrunc quation has a non-trivial solution for all larg rims Thorm 1. Lt n > 1. Thn thr xists an intgr S(n) such that for all rims > S(n) th congrunc x n + y n z n (mod ) has a solution in th intgrs, such that dos not divid xyz. Th condition dos not divid xyz is to avoid trivial solutions of th congrunc, such as x y z 0 or x 0, y z (mod ). Th abov rsult follows from a smingly vry diffrnt on. Thorm 2. Lt r 1. Thn thr is a natural numbr S(r), such as if N S(r) and if th numbrs {1, 2,..., N} ar colord with r colors, thn thr ar thr of thm x, y, z of th sam color satisfying th quation: x + y = z. W ll rfr such numbrs x, y, z as a monochromatic Schur tril, and S(r) is dfind to b th smallst natural numbr for which th abov statmnt holds. Th roof is basd on Ramsy s thorm for coloring of th dgs of comlt grah on N vrtics, which you hav rad in th first chatr of th txtbook. Proof. Lt c : [1, N] [1, r] b an r- coloring of th first N intgrs. Dfin a corrsonding coloring of th comlt grah with vrtics 1, 2,..., N by coloring th dg (i, j) to c( i j ). By Ramsy s thorm if N R(r, 3) thn thr is a monochromatic triangl. If i < j < k ar th vrtics of this triangl, listd in incrasing ordr, thn writing x = j i, y = k j and z = k i w hav that c(x) = c(y) = c(z) and x + y = z, thus thy form a monochromatic Schur tril. To rov Thorm 1, now rquirs only a bit of lmntary algbra. Lt b a rim, and lt Z dnot th congrunc classs modulo. Ths congrunc classs can b addd 1
2 REU SUMMER 2005 and multilid togthr (sums and roducts of congrunt numbrs stays congrunt), and Z bcoms a fild undr ths orations. In articular Z = Z {0} is a grou undr multilication. For givn n, lt G n = {x n : x Z }. Thn G n is a subgrou of Z, thus thr is a st of lmnts a 1, a 2,..., a r, such that Z is artitiond as Z = a 1 G n a 2 G n... a r G n If you don t know this, just notic that th rlation: x y if x = ay for som a G n is an quivalnc rlation, and a i G n ar th quivalnc classs. Also r is th numbr of x Z such that x n = 1 in Z (that is x n 1 (mod )). This follows from th fact that x x n is a homomorhism of Z. So r is th numbr of roots of th olynomial x n 1 in th fild Z, hnc r n. Now, lt c(x) = i if x a i G n, which givs an r coloring of Z = {1, 2,..., 1}. If 1 S(n) S(r) thn thr is a monochromatic Schur tril, that is thr ar intgrs x, y, z, non of which is divisibl by, such that sinc a i is not divisibl by it follows This rovs Thorm 1. a i x n + a i y n a i z n (mod ) x n + y n z n (mod ) 2. Fourir analytic aroach In this sction w giv an altrnativ roof of Thorm 1, basd urly on Fourir analysis on th additiv grou Z. Lt ω = 2πi sum: dnot th -th root of unity, and for k Z considr th so-calld xonntial S k = x Z ω kxn = 1 x=0 kxn 2πi Lt us mak som siml obsrvations first, whos roof is lft as an xrcis. (i) If a Z, a 0 thn S k = S ka n. Indd by making a chang of variabls: x ax, S k transforms into S ka n. (ii) Thr is a basic orthogonality trick, oftn usd in Fourir analysis: 1 k=0 { 2πi km if m 0 (mod ) = 0 othrwis
Us this to show that: k Z S k 2 = SCHUR S THEOREM 3 2πi k(x n y n ) x,y Z k Z = N whr N = {x, y Z : x n = y n }., that is th numbr of solutions of th quation x n = y n within th fild Z. (iii) Show that N 1 + n. Th ida is if x 0 thn writing y = ux whr u n = 1. (iv) Lt G n = {a n : a Z }, thn G n ( 1)/n. Lmma 1. If k Z, k 0 thn on has th stimat x Z 2πi kxn 2n 1 2 Proof. Using th abov obsrvations G n S k 2 = S ka n 2 S l 2 n 2 a n G n Thus This rovs th lmma. l Z S k 2 n2 2 1 2n2 Proof of Thorm 1. Lt M dnot th numbr of ordrd trils x, y, z in Z satisfying: x n + y n = z n. Thn using th orthogonality trick again, on can gt an analytic xrssion for M: M = 1 1 k(x n +y n z n ) 2πi x,y,z Z k=0 = 1 1 k=0 S 2 k S k whr S dnots th comlx conjugat. Th scond inquality is obtaind by taking th summation in x, y and z first. Th dominating trm in th abov sum is S 0 =, indd M 2 1 1 k=1 S k 3 2 (2n 2 ) 3 2 3 2 1 2 2 if th rim is larg nough: 16n 6. Finally w hav to count th solutions whr x, y or z is th zro lmnt of Z. Th numbr of such solutions is at most: 1 + 3n < 1 2 2. Hnc thr must b a solution x, y, z in Z such that xyz 0. This rovs th thorm. Not that th Fourir analytic roof givs a much smallr bound 16n 6 thn th combinatorial on 3n!, on th rim for larg valus of n. W ll s that th Ramsy numbr R(3, n) is xonntially larg, so th bound on cannot b ssntially rducd in th first argumnt. This hnomnon coms u in many diffrnt contxt; whnvr Fourir analysis can b usd it oftn lads to much bttr bounds than combinatorics.
4 REU SUMMER 2005 3. Exrciss 1. Prov that R(3, r) 3r! by using induction on r. 2. Prov that S(r) 3r +1 2 by comlting th following sts; (i) S(2) = 5 (ii) If c is an r coloring of [1, N] such that thr is no monochromatic Schur tril, thn dfin an r + 1 coloring of [1, 3N + 1] th following way. Color th two blocks [1,N] and [2N+2,3N+1] th sam way as original coloring, and color ach numbr in [N+1,2N+1] by a nw color. Show that thr is no monochromatic Schur tril in th nw coloring. Conclud that S(r + 1) 3S(r) 1, and do induction on r. Not: A mor dtaild dscrition of ths sts ar in th txtbook, but it is usful to try it by yourslf first. 3. Show that if th dgs comlt grah on N vrtics ar colord with 2 colors, thn thr ar at last n 3 /24 monochromatic triangls, and this is th bst ossibl lowr bound. Hint: Estimat th numbr of 2-colord triangls from abov. Notic that for a 2-colord triangl, thr ar xactly 2 vrtics with diffrnt colord dgs. 4. Show that for any 2-coloring of th first N intgrs, thr ar at last n 2 /48 Schur trils. Us xrcis 1. Not: In th txtbook, you find th bst lowr bound n 2 /22 for th numbr of Schur trils in a 2-coloring of [1, N]. It is worth to rad th argumnt, aftr you comltd th rvious xrcis. 5. Prov that for any air of intgrs r and n, thr is a P (r, n), such that if P (r, n) is a rim, and Z is colord with r colors, thn thr is a monochromatic tril x, y, z in Z such that x n + y n = z n and xyz 0 in th fild Z. That is thr xist monochromatic Frmat trils mod. 6. Lt n b givn. Prov that if q is such that all rim divisors of q ar largr thn S(n), thn thr xists a tril x, y, z such that xyz is rlativ rim to q and thy satisfy: x n + y n z n (mod q) by comlting th following sts: (i) Suos x, y, z solvs: x n + y n z n (mod r ) such that xyz. Show that thr xist a tril u, v, w such that X = r u + x, Y = r v + y and Z = r w + z solvs: X n + Y n Z n (mod r+1 )
SCHUR S THEOREM 5 (ii) Lt q 1, q 2 b such that (q 1, q 2 ) = 1, that is rlativ rims. Show that to any air of congrunc classs x 1 (mod q 1 ) and x 2 (mod q 2 ), thr is a uniqu rsidu class x (mod q 1 q 2 ) such that: x x 1 (mod q 1 ) and x x 2 (mod q 2 ). Show that to vry air of Frmat trils (mod q 1 ) and (mod q 2 ), thr corrsonds a Frmat tril (mod q 1 q 2 ). 7*. Lt 0 < δ < 1, and n a ositiv intgr. Show that thr xists an N(δ, n) such that if > N(δ, n) is a rim, and A Z is a st of at last δ lmnts, thn thr xits a tril x, y, z such that x A, y A, z Z and x n + y n = z n in Z. This strngthns Schur s thorm, on can find a Frmat tril such that two lmnts is in a givn st A of ositiv dnsity δ. S th xlanation of th notion of dnsity blow. Hint: Lt A = {x n : x A}. Show that A δ. Now it is nough to find a n solution of: x + y = z n whr x, y A. Lt N dnot th numbr of such trils x, y, z in Z. Find th following analytic xrssion for N: N = 1 χ A (k) 2 S k k Z whr χ A (k) = xk 2πi x A is th Fourir transform of th indicator function χ A of th st A, and S k is th xonntial sum dfind abov. Show that th main trm in th sum is coms from k = 0 by using Planchrl s thorm in Z. 8.** Is it tru that for any air of intgrs r and n, thr is a Q(r, n), such that if Q(r, n) is a rim, and if A Z is a st of at last /r lmnts, thn thr ar thr lmnts x, y, z of A such that x n + y n = z n and xyz 0 in Z? Not: Problm 8 is th dnsity vrsion of Exrcis 5. If A is a subst of Z = {1, 2,..., } thn th ratio A / is calld th dnsity of th st A. If you color th st {1, 2,..., } with r colors, or in othr words if you artition it into r substs, thn on of thm has to hav dnsity at last 1/r. If you r looking for substs with crtain nic rorty (such as Frmat trils, arithmtic rogrssions, tc.) thn you can ask both th coloring or th dnsity qustion; whthr you find a monochromatic such subst, or whthr vry subst A of dnsity at last 1/r contains such a subst. By th rmark abov th dnsity rsult always imlis th coloring on, and oftn is considrably hardr. W ll s a famous xaml, Roth thorm which is th dnsity vrsion of Van dr Wardn s rsult for arithmtic 3-rogrssions.