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The Erd˝Os-Heilbronn Conjecture Math 21 − 801 (Additive Combinatorics) Carnegie Mellon University

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The Erd˝Os-Heilbronn Conjecture Math 21 − 801 (Additive Combinatorics) Carnegie Mellon University The Erdos-Heilbronn} Conjecture Math 21 ¡ 801 (Additive Combinatorics) Carnegie Mellon University - Spring 2009 1. Motivation The problems we will be considering lie in the area of Additive Number Theory. This relatively young area of Mathematics is part of Combina- torial Number Theory and can best be described as the study of sums of sets of integers. As such, we begin by stating the following de¯nition: De¯nition 1.1. [Sumset] For sets A and B (usually subsets of Z=pZ), de¯ne A + B := fa + b j a 2 A; b 2 Bg: A simple example of a problem in Additive Number Theory is given two subsets A and B of a set of integers, what facts can we determine about A + B? Note that a very familiar result in Number Theory, namely Lagrange's theorem that every nonnegative integer can be written as the sum of four squares, can be expressed in terms of sumsets. In particular, if we let N0 be the set of nonnegative integers and if we let S be the set of all integers that are perfect squares, then Lagrange's Four Square Theorem has the form Theorem 1.2. [Lagrange's Four Square Theorem] N0 = S + S + S + S 2 where N0 = fx 2 Z j x ¸ 0g and S = fx j x 2 Zg. As well the binary version of Goldbach's Conjecture can be restated in terms of sumsets. In particular, Conjecture 1.3. [Goldbach's Conjecture] Let E = f2x j x 2 Z; x ¸ 2g and let P = fp 2 Z j p is prime g. Then E ⊆ P + P: (1) In other words, every even integer that is greater than 2 is the sum of two primes. Notice that we do not have set equality in equation (1) because 2 2 P. Once again, 2 is the \odd" prime. 1 Additive Combinatorics J P Wheeler 2. The Problems We Consider 2.1. The Cauchy-Davenport Theorem. The ¯rst result we will be concerned with is a theorem proved by Cauchy1 in 1813 [6] and independently by Davenport in 1935 [8] (Dav- enport discovered in 1947 [9] that Cauchy had previously proved the theorem). In particular, Theorem 2.1. [Cauchy-Davenport] Let A and B be nonempty subsets of Z=pZ with p prime. Then jA + Bj ¸ minfp; jAj + jBj ¡ 1g where A + B := fa + b j a 2 A and b 2 Bg: We note that in 1935 Inder Cholwa [7] extended the result to composite moduli m when 0 2 B and the other members of B are relatively prime to m. 2.2. The Erd}os-HeilbronnConjecture. The second result we consider is a slightly di®erent version of the ¯rst. In the early 1960's, Paul Erd}osand Hans Heilbronn conjectured that if the addition in the Cauchy-Davenport Theorem is restricted to distinct elements the lower bound slightly changes. Erd}osstated this conjec- ture in 1963 during a number theory conference at the University of Colorado [11]. Interestingly, Erd}osand Heilbronn did not mention the conjecture in their 1964 paper on sums of sets of congruence classes [14] though Erd}osmentioned it often in his lectures (see [15], page 106). Eventually the conjecture was formally stated in Erd}os'contribution to a 1971 text [12] as well as in a book by Erd}osand Graham in 1980 [13]. In particular, Theorem 2.2. [Erd}os-Heilbronn Conjecture] Let p be a prime and A, B ⊆ Z=pZ with A 6= ; and B 6= ;. Then jA+_ Bj ¸ minfp; jAj + jBj ¡ 3g, where A+_ B := fa + b mod p j a 2 A, b 2 B and a 6= b g. The conjecture was ¯rst proved for the case A = B by J.A. Dias da Silva and Y.O. Hamidounne in 1994 [10] using methods from linear 1Cauchy used this theorem to prove that Ax2 +By2 +C ´ 0(mod p) has solutions provided that ABC 6´ 0. This is interesting in that Lagrange used this result to establish his four squares theorem. 2 Additive Combinatorics J P Wheeler algebra with the more general case established by Noga Alon, Melvin B. Nathanson, and Imre Z. Ruzsa using the polynomial method in 1995 [1]. 3. Preliminary Matter The following fact from ¯eld theory is essential to our work. Theorem 3.1. Let F be a ¯eld and suppose p(x) 2 F[x] where degree p(x) = d. If p(x) is not the zero polynomial, then p(x) can have at most d distinct roots in F. We use this to establish the following Lemma which is fundamental to the Polynomial Method. Lemma 3.2 (Alon-Tarsi [3]). Let f(x; y) be a polynomial with coe±cients in an arbitrary ¯eld F and of degree at most k ¡ 1 in x and degree at most l ¡ 1 in y. Let A and B be subsets of F with jAj = k and jBj = l. If f(a; b) = 0 for all a 2 A and for all b 2 B, then f(x; y) ´ 0. Proof. We have Xk¡1 Xl¡1 i j f(x; y) = fi;jx y : i=0 j=0 Grouping together terms of degree xi and factoring enable us to write Xk¡1 i f(x; y) = gi(y)x i=0 Pl¡1 j where gi(y) = j=0 fi;jy for each i = 0; : : : ; k ¡ 1. As well, ¯x b 2 B and put Xk¡1 i h(x) = gi(b)x : i=0 Then for all a 2 A, h(a) = f(a; b) = 0. But jAj = k while degree h(x) · k ¡ 1. Hence by Theorem 3:1, h(x) ´ 0 giving us gi(b) = 0 for all i. This is true for each b 2 B. Thus, since jBj = l and degree gi(y) = l ¡ 1, Theorem 3:1 again gives gi(y) ´ 0 for each i. Hence we have f(x; y) ´ 0. ¤ 3 Additive Combinatorics J P Wheeler We will also need Lemma 3.3 (Alon-Nathanson-Ruzsa [1]). Let A be a ¯nite subset of an arbitrary ¯eld F with jAj = k. Then for every r ¸ k there exists a polynomial gr(x) 2 F[x] of degree at most r k ¡ 1 such that gr(a) = a for all a 2 A. Proof. Fix r ¸ k and let A = fa1; : : : ; akg. Our goal is to construct the appropriate polynomial C(x) of degree at most k ¡ 1. Put C(x) = k¡1 c0 + c1x + ¢ ¢ ¢ + ck¡1x . Thus we need k¡1 r C(a1) = c0 + c1 ¢ a1 + ¢ ¢ ¢ ck¡1a1 = a1 k¡1 r C(a2) = c0 + c1 ¢ a2 + ¢ ¢ ¢ ck¡1a2 = a2 . k¡1 r C(ak) = c0 + c1 ¢ ak + ¢ ¢ ¢ ck¡1ak = ak: Note that this gives rise to a k by k matrix and, by Cramer's Rule, this matrix has a solution if the determinant of the coe±cient matrix is nonzero. But this matrix is just Y V (a1; : : : ; an) = (aj ¡ ai) 1·i<j·n which is not 0. ¤ This proof is the one provided by Alon, Nathanson, and Ruzsa. A much simpler means of establishing the result is by making use of Lagrange Interpolation: Lagrange Interpolation. Fix r ¸ k and let A = fa1; : : : ; akg. Put 0 1 X B Y C B r x ¡ aj C gr(x) := @ai A : ai ¡ aj ai2A aj 2A j6=i r Then gr(ai) = ai for all i, 1 · i · k. ¤ We note that Lemma 3.2 was originally stated in [3] for a polynomial with coe±cients in Z where A and B are subsets of Z. The result was proven for arbitrary ¯elds in [1]. 4 Additive Combinatorics J P Wheeler 4. The Method Employed We ¯rst prove Theorem 4.1. Suppose A and B are nonempty subsets of Fp = Z=pZ where p is a prime. Further suppose that jAj = k, jBj = l, and that k 6= l. Then jA+_ Bj ¸ minfp; k + l ¡ 2g where A+_ B := fa + b mod p j a 2 A, b 2 B and a 6= b g. Proof. Without loss of generality we may assume that 1 · l = jBj < k = jAj · p: Now if k + l ¡ 2 > p, let l0 = p ¡ k + 2. Then 2 · l0 · l < k and k + l0 ¡ 2 = p: Choose B0 ⊆ B such that jB0j = l0. Hence, if the theorem holds for the sets A and B0, then jA+_ Bj ¸ jA+_ B0j ¸ k + l0 ¡ 2 = p = minfp; k + l ¡ 2g: Therefore we may assume that k + l ¡ 2 · p: (2) We form the set C = A+_ B and assume for contradiction that jCj · k + l ¡ 3 (3) and we will put m = k + l ¡ 3 ¡ jCj: (4) Thus, by (3), m is nonnegative. We form a polynomial in Fp[x; y] by de¯ning Y f0(x; y) := (x + y ¡ c): (5) c2C Hence deg(f0) = jCj · k + l ¡ 3 (6) 5 Additive Combinatorics J P Wheeler where deg(f0) is the homogeneous degree of f0. Also f0(a; b) = 0 for all a 2 A; b 2 B; a 6= b: (7) As well de¯ne Y f1(x; y) = (x ¡ y)f0(x; y) = (x ¡ y) (x + y ¡ c): (8) c2C Then deg(f1) = 1 + jCj · k + l ¡ 2 (9) and f1(a; b) = 0 for all a 2 A; b 2 B: (10) Lastly we form the polynomial Y m m f(x; y) = (x + y) f1(x; y) = (x + y) (x ¡ y) (x + y ¡ c): (11) c2C Note that deg(f) = m + 1 + jCj = k + l ¡ 2 (12) and that f(a; b) = 0 for all a 2 A; b 2 B: (13) Since Y f(x; y) = (x ¡ y)(x + y)m ((x + y) ¡ c) c2C = (x ¡ y)(x + y)m+jCj + lower order terms; we have X i j k+l¡3 f(x; y) = fi;jx y = (x ¡ y)(x + y) + lower order terms: i;j¸0 i+j·k+l¡2 6 Additive Combinatorics J P Wheeler By assumption, p ¸ k + l ¡ 2, and we have k; l 6= 0.
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