On Egyptian fractions
aa r X i v : . [ m a t h . N T ] A p r ON EGYPTIAN FRACTIONS
MANUEL BELLO HERN ´ANDEZ , MANUEL BENITO , AND EMILIO FERN ´ANDEZ
Abstract.
We find a polynomial in three variables whose values at nonnega-tive integers satisfy the Erd˝os-Straus Conjecture. Although the perfect squaresare not covered by these values, it allows us to prove that there are arbitrarilylong sequence of consecutive numbers satisfying the Erd˝os-Straus Conjecture.We conjecture that the values of this polynomial include all the prime numbersof the form 4 q +5, which is checked up to 10 . A greedy-type algorithm to findan Erd˝os-Straus decomposition is also given; the convergence of this algorithmis proved for a wide class of numbers. Combining this algorithm with the men-tioned polynomial we verify that all the natural numbers n , 2 ≤ n ≤ × ,satisfy the Ed˝os-Straus Conjecture. Introduction
One of the most famous conjecture on Egyptian fractions is the Erd˝os-StrausConjecture (ESC):
Given a positive integer n ≥ there exist positive integers ( x, y, z ) ∈ Z > such that (1) 4 n = 1 x + 1 y + 1 z . In this case we say that n is an Erd˝os-Straus’ number and refer to (1) as an Erd˝os-Straus decomposition of n . Sierpi´nski and Schinzel extent to a more general conje-ture replacing 4 in (1) by other fixed positive integer m ≥
4. The roots of these liein the minimum number of Egyptian fractions needed to decompose a fraction assum of Egyptian fractions (see, for example, [2], [4], [11], [13], [16] and [18]). ESChas been verified for all integer up to a bound by many authors: Straus, Bernstein[1], Yamamoto [19], Swett [17], etc. Swett has checked ESC for all n ≤ . If n isan Erd¨os-Straus’ number, then it also holds true for integers which are divisible by n ; moreover, q +3 = q +1 + q +1)(4 q +3) and if we have the factorization n = a · b · c , n = a ( a + b ) c + b ( a + b ) c . Therefore, ESC should be only checked for primes n = 4 q + 1.Because(2) 4 abc − a − b = 1 a bc − + 1 a ( ac − bc − + 1( ac − bc − n , when bc ≡ n = abc − a − b are Erd˝os-Straus’ numbers.Moreover, if we replace the condition bc ≡ bc ≡ m ), then theSierpi´nski and Schinzel conjecture holds for n = abc − a − b . Since we are interested Mathematics Subject Classification.
Primary 11D68; Secondary 11A07,11N32.
Key words and phrases.
Egyptian fraction, Erd˝os-Straus Conjecture.This research was supported in part from ‘Ministerio de Ciencia y Tecnolog´ıa’, ProjectMTM2009-14668-C02-02. in n = 4 q + 1, let n = p ( α, β, γ ) be the polynomial p : Z ≥ → Z given by(3) p ( α, β, γ ) = ( α + 1)(4 β + 3)(4 γ + 3) − ( α + 1) − (4 β + 3) . This parametric solution of ESC let us to prove the following result:
Theorem 1.
There are arbitrarily long sequence of consecutive numbers satisfyingErd˝os-Straus Conjecture.
We have checked that the set N = (cid:8) n ∈ Z > : ∃ ( α, β, γ ) ∈ Z ≥ , n = p ( α, β, γ ) (cid:9) contains all the prime numbers of the form n = 4 q + 5 , q ∈ Z ≥ , n < . Weprove, see Lemma 7 below, that N does not contain the perfect squares.In Section 2 we list several relations for ESC. One interesting conclusion in thissection is given in Lemma 3: it is impossible to generate “naturally” a finite numberof congruence classes satisfying ESC which contain all the prime. Section 3 containsa proof of Theorem 1. In Section 4 we prove that the set N does not contain theperfect squares and several other additional remarks on ESC. We describe a fastalgorithm to construct a decomposition of n as sum of three Egyptian fractionsand we prove the convergence of that algorithm for a large class of numbers in thelast Section 5. 2. Auxiliary relations
The Erd˝os-Straus Conjecture shows an arithmetic property of the natural num-bers (see Lemma 6 below). We can split any fraction n as sum of two Egyptianfraction but the same does not hold for n for all n ∈ Z > . It is very well knownthat the equation n = x + y has solutions ( x, y ) ∈ Z > if and only if n has a divisor m with either m ≡ m ≡ n Lemma 1.
Let n be a prime; n is an Erd˝os-Straus’ number if and only if thereexists ( a, b, c, d ) ∈ Z > such that some of the following conditions holds: (4 abc − d = ( a + b ) n, (4) (4 abc − d = an + b. (5) Proof.
If (4 abc − d = ( a + b ) n or (4 abc − d = an + b holds, dividing theseequations by abcdn , we have respectively4 n = 1 abcn + 1 bcd + 1 acd , (6) 4 n = 1 abcn + 1 bcd + 1 acdn . (7)On the other hand, if (1) holds, then 4 xyz = n ( xy + yz + zx ) . Since n is prime, n divides x, y or z . Of course, n does not divide all the three numbers x, y, z becausethis trivially yields the contradiction 4 = x/n + y/n + z/n with x/n, y/n, z/n positiveintegers. Hence, we can assume without lost of generality x = an . Thus (1) isequivalent to(8) 4 a − na = 1 y + 1 z ⇔ na = 1(4 a − y + 1(4 a − z . N EGYPTIAN FRACTIONS 3
Since n is prime and (4 a − , a ) = 1 , we have two cases: (4 a − , n ) = 1, and(4 a − , n ) = n .In the first case, there exist a , a , a ∈ Z + such that a = a a a , ( na , a ) = 1,and(9) (4 a − y = na ( na + a ) a , (4 a − z = a ( na + a ) a . Because ( na, a −
1) = ( na a a , a −
1) = 1, there exist α, β such that y = αna a , z = βa a . From (9), we have (4 a − α = ( na + a ) = (4 a − β, it means α = β . Therefore, A = a , B = a , C = a , and D = α satisfy(4 ABC − D = ( nA + B ) . Now we consider the second case (4 a − , n ) = n . So, there exists j such that(10) 4 a − jn. Setting this expression in (8), we have a = jy + jz . Thus, there exist a , a , a ∈ Z + such that a = a a a , ( a , a ) = 1 and(11) jy = a ( a + a ) a , jz = a ( a + a ) a . Since ( j, a ) = ( j, a a a ) = 1, there exist α, β such that y = αa a , z = βa a .Setting these expression in (11), we have jα = a + a = jβ, it is α = β . Multiplyingin (10) by α , we obtain A = a , B = a , C = a , D = α such that(4 ABC − D = ( A + B ) n. (cid:3) The relations (4) and (5) are equivalent to (6) and (7), respectively. Since in thefirst decomposition of n , n divides one denominator but is coprime to the others,while in the second it is only coprime to one of them, these are respectively refereedas Type I and Type II decomposition of n (see [2]). Observe that it is not possiblefor all three of x, y, z in (1) to be divisible by n .Changing variables in (4) and (5) the following lemma follows. Lemma 2. (1)
Let n be a prime. There exist positive integers a, b, c, d suchthat (4) holds if and only if there are positive integers α, β, γ, δ such that (12) δn = (4 αβγδ − − α γ. (2) Let n ∈ Z , n ≥ . There exist positive integers a, b, c, d such that (5) holdsif and only if there are positive integers α, β, γ, δ satisfying (13) n = (4 αβγ − δ − β γ. Proof. (1) If n is a prime and (4) holds, then d divides a + b . Set e = a + bd ;hence, b = de − a and (4) yields en = (4 acde − − a c. Setting δ = e , α = a , β = d , and γ = c , we obtain (12).(2) The relation (5) is equivalent to b + d is divisible by a and n + s = 4 bcd ,where s = b + da ⇔ d = as − b . Setting α = a, β = b, γ = c, δ = s we obtainimmediately (13). (cid:3) Equation (12) implies that δn is an Erd˝os-Straus’ number with Type I and TypeII decomposition, but it is trivial because if n ≡ δ ≡ − β = γ = 1 in(13), it follows that if n + 4 has a divisor congruent to 3 modulo 4, then n is an M. BELLO-HERN´ANDEZ, M. BENITO, AND E. FERN´ANDEZ
Erd˝os-Straus’ number. So by the Landau prime ideal theorem (see [9] or [12] pp.266-268) the set of positive integers which are no Erd˝os-Straus’ numbers has zerodensity. A sharp estimate on the density of Erd˝os-Straus’ numbers can be found in[18] (see also [2], [6], [7], and [8]).In [19] Yamamoto (see also [15]) proves that a perfect square n does not satisfyneither (4) nor (5) with some restriction in the parameters (see the paragraphsat the begging of Subsection 4.2 below), so fixing the parameters a, b, c, d in theseequations we can not generated a complete residue system. The following lemmaemphasis in this remark without use that result of Yamamoto. Lemma 3.
We can not generate a finite number of congruence classes containingall the primes of the form q + 5 fixing in (4) or (5) three of the four parametersin a finite subset of Z > and the remain parameter free in Z > .Proof. Let S a = { ( b, c, d ) ∈ Z > } denote a finite subset of Z > of values takenby ( b, c, d ). From (4) or (5), when a is free taking any positive number, eachfixed vector ( b, c, d ) ∈ S a may generate a residue class of Erd˝os-Straus’ numbers n .Equivalently we define S b , S c and S d . We fixe four such subset S a , S b , S c , S d andfirst prove that there exist infinite prime numbers n which can not be generated by(5) with parameters ( a, b, c, d ) with three of them in the corresponding set S a , S b , S c , or S d .The numbers n generate by (5) with ( b, c, d ) ∈ S a are a finite set because a | ( b + d ).Moreover, observe that in the proof of (13) in Lemma 2, the numbers n generateby (5) or equivalently by (13) with ( b, c, d ) ∈ S c are given by n = 4( αδ − β ) βγ − δ = 4 bcd − b + da , ( a, b, d ) ∈ S c , where α = a , β = b , γ = c , and δ = b + da . Taking T c = lcm { bd : ( a, b, d ) ∈ S c } , the numbers in { T c t + 1 : t ∈ Z > } (in particular the prime numbers in { T c t + 1 : t ∈ Z > } ) can not be generated by (13) with ( a, b, d ) ∈ S c , herelcm { bd : ( a, b, d ) ∈ S c } denotes the least common multiple of the numbers bd when( a, b, d ) runs through the set S c . Suppose for contradiction that for a given t ∈ Z > there exist ( a , b , d ) ∈ S c and c ∈ Z > such that4 T c t + 1 = 4 b cd − b + d a ⇔ b d ( c − e ) − b + d a , where e ∈ Z > , but it is a contradiction since 1 is not an Erd˝os-Straus’ number.Using the same arguments and notations as before, when ( a, b, c ) ∈ S d , thenaccording to the proof of (13) in Lemma 2, ( α, β, γ ) lies in a finite subset S δ of Z > , and δ ∈ Z > . Taking T d = lcm { (4 αβγ −
1) : ( α, β, γ ) ∈ S δ } , the numbers in { T d t + 1 : t ∈ Z > } can not be generated by (13) with ( a, b, c ) ∈ S d . Suppose fora t ∈ Z > there exists ( α , β , γ ) ∈ S δ and δ ∈ Z > such that4 T d t + 1 = (4 α β γ − δ − β γ ⇔ α β γ − δ − e ) − β γ for some e , again it is not possible because 1 is not an Erd˝os-Straus’ number. Fromthe symmetry of (13) in b and d the same conclusion holds for S b for numbers { T b t + 1 : t ∈ Z > } , where T b is defined as T d . Of course all the numbers { T b T c T d t + 1 : t ∈ Z > } can not be generated by parameters ( a, b, c, d ) with three of them in the correspond-ing set S a , S b , S c or S d . N EGYPTIAN FRACTIONS 5
The same arguments work for (4). For example, one of the parameters a or b in(4) can not take values in an infinite subset of positive integers while other threebelong to a finite subset. In fact, observe that (4) is symmetric in a and b , andaccording to the proof of (12) in Lemma 2, the number e = a + bd divides 1 + 4 a c ,so that if ( a, c, d ) runs over a finite set, the numbers of divisors of 1 + 4 a c is finite.Moreover, the numbers n generate by (4) or equivalently by (12) with ( b, c, d ) ∈ S c do not contain { T ′ c t + 1 : t ∈ Z ≥ } where T ′ c = lcm { abd : ( a, b, d ) ∈ S c } . Supposefor contradiction that for a given t there exist ( a , b , d ) ∈ S c and c ∈ Z > suchthat ( a + b )(4 T ′ c t + 1) = (4 a b c − d ⇔ ( a + b )1 = (4 a b ( c − e ) − d where e = ( a + b ) T ′ c ta b d , but it is a contradiction since 1 is not an Erd˝os-Straus’number. For ( a, b, c ) ∈ S d , we can not generate the numbers 4 T ′ d t + 1, where T ′ d = lcm { (4 abc −
1) : ( a, b, c ) ∈ S d } .Therefore, equations (4) and (5) can not generate all the numbers(14) { T b T c T d T ′ c T ′ d t + 1 : t ∈ Z > } except possibly a finite number of them with three parameters of a, b, c, d in thecorresponding set S a , S b , S c or S d . (cid:3) Remark . Note that using Yamamoto result we can actually prove that equations(4) and (5) can not generate all the numbers { T b T c T d T ′ c T ′ d t + n : t ∈ Z > } except possibly a finite number of them with three parameters of a, b, c, d in thecorresponding finite set S a , S b , S c or S d , where n is a quadratic residue modulo4 T b T c T d T ′ c T ′ d .Many papers devote special attention to Type II decomposition since parametricsolutions of ESC are easy obtained for this case, see (13). Our following lemmaallows us to find a parametric solution (3) of Type I decomposition. Lemma 4.
Let n ∈ Z > . There exist positive integers a, b, c, d such that (4) holdsif and only if there are x, t, λ such that (15) (cid:26) x n + tλ ∈ Z > , n + λ x t ∈ Z > are satisfied.Proof. Assume that there exist positive integers x, t, λ, y, z such that x n + tλ = y, n + λ x t = z ⇔ x n + t = y λ, λ = 4 z x t − n ⇔ ( x + y ) n = (4 x y z − t. Therefore, taking a = x, b = y, c = z, d = t we have (4 abc − d = ( a + b ) n . (cid:3) Remark . If n is an Erd˝os-Straus’ number and (15) is satisfied for positiveintegers x, t, λ , then for all j ∈ Z ≥ , N = n + 4 xtλ j is also an Erd˝os-Straus’number. In fact, ( x N + tλ = x ( n +4 xtλ j )+ tλ = xn + tλ + 4 x tj ∈ Z + N + λ x t = ( n +4 xtλ j )+ λ x t = n + λ x t + λ j ∈ Z + M. BELLO-HERN´ANDEZ, M. BENITO, AND E. FERN´ANDEZ In particular, the set of Erd˝os-Straus’ numbers is an open set in Furstenberg’stopology (see [14], p. 34).
Remark . The values n = p ( α, β, γ ) = ( α + 1)(4 β + 3)(4 γ + 3) − ( α + 1) − (4 β + 3)satisfy (15) with x = 1, t = α + 1, and λ = 4 β + 3.The parametric relation (3) for p = 4 q + 5 is useful to rewrite for q . Since p ( α, β, γ ) = 4 q ( α, β, γ ) + 5 = 4 (((4 β + 3) γ + (3 β + 2))( α + 1) − ( β + 2)) + 5 , the following lemma follows: Lemma 5.
If there exists ( α, β, γ ) ∈ Z ≥ such that (16) q = q ( α, β, γ ) = ((4 β + 3) γ + (3 β + 2))( α + 1) − ( β + 2) , then the Erd˝os-Straus Conjecture holds for p = 4 q + 5 . Consecutive numbers satisfying ESC
Next, using the Chinese Remainder Theorem and Lemma 5 we prove Theorem 1.Actually, we prove that there exist arbitrarily long sequence of consecutive residuesclass such that n has Type I decomposition for all n in this residues class. Proof of Theorem 1.
Of course, to get Theorem 1 it is enough to consider“consecutive” numbers n ≡ N be an arbitrary positive integer and let A be a subset of Z > containing N consecutive nonnegative integers, for example A = { , , . . . , N − } . If { β , β } ⊂ A , then the greatest common divisor of 4 β + 3and 4 β + 3, (4 β + 3 , β + 3), is an odd number and(4 β + 3 , β + 3) | (4 β + 3 − (4 β + 3)) = 4( β − β ) ⇒ (4 β + 3 , β + 3) | β − β ) = 3 β + 2 − (3 β + 2) . Thus, by the Chinese Remainder Theorem, there is a natural number T such that T ≡ β j + 2 (mod 4 β j + 3) , ∀ β j ∈ A, i.e., there exist positive integers γ j such that T = (4 β j + 3) γ j + 3 β j + 2 , ∀ β j ∈ A. According to Lemma 5 all n = 4 q + 5 with q in the consecutive residue class − ( β j + 2) (mod T ), β j ∈ A , satisfy the Erd˝os-Straus conjecture. (cid:3) Remark . Moreover, we can also prove that there exists arbitrarily long sequenceof consecutive numbers n such that n has Type II decomposition. Let A be a setof natural numbers containing consecutive numbers, for each a ∈ A , we choosenatural numbers β ( a ) and γ ( a ) such that a = β ( a ) γ ( a ) (such representation isunique taking γ ( a ) free of squares). We chose T as the least common multiple of(4 β ( a ) γ ( a ) − a takes the values in A , i.e. T = lcm { (4 β ( a ) γ ( a ) −
1) : a ∈ A } , and δ = (cid:26) , if T ≡ , , if T ≡ − , then, using (13) in Lemma 2, it follows that all the fractions T δ − a , a ∈ A, haveType II decomposition. N EGYPTIAN FRACTIONS 7 Additional remarks q − Conjecture.Lemma 6.
ESC is true if and only if for each q ∈ Z > there exists ( x, y, z ) ∈ Z ≥ such that one of the following relations holds q = 1 + 3 x + 3 y + 4 xy, (17) q = 5 + 5 x + 5 y + 4 xy, (18) q = q ( x, y, z ) . (19) where q ( x, y, z ) is defined by (16).Proof. Because q = 1 + 3 x + 3 y + 4 xy ⇔ q + 5 = (4 x + 3)(4 y + 3) ,q = 5 + 5 x + 5 y + 4 xy ⇔ q + 5 = (4( x + 1) + 1)(4( y + 1) + 1) , all q for composite numbers 4 q +5 satisfy (17) or (18). We also have for q = q ( x, y, z )define by (16), q = q ( x, y, z ) = p ( x, y, z ) − . Hence if all q ∈ Z > satisfies (17), (18) or (19), the value of q for all prime numbersof the form 4 q + 5 lies in N and the proof is concluded using Lemma 5. (cid:3) Remark . We have verified that N contains all primes of the form n = 4 q + 5 < . To this aim we have generated the classes of equivalence of numbers containedin N translating t units each variable; i.e. N x = p ( x, y, z ) + f ( y, z ) t def = p ( x, y, z ) + 4 (4 y + 3)(4 z + 3) − t,N y = p ( x, y, z ) + f ( x, z ) t def = p ( x, y, z ) + 4(( x + 1)(4 z + 3) − t,N z = p ( x, y, z ) + f ( x, y ) t def = p ( x, y, z ) + 4( x + 1)(4 y + 3) t. Starting with x, y, z ∈ { , } we obtain the following equivalence class5 + 8 t, t,
13 + 20 t,
17 + 20 t,
13 + 28 t,
37 + 52 t, t ∈ Z ≥ , then we sieve prime numbers congruent to 1 modulo 4 in these congruence classes.We are checked by computer calculations that the remaining primes < belongto N . Conjecture.
We conjecture that all the natural number q ∈ Z > can be writ-ten as one of the three above relations (17), (18) or (19). We refer it as the q − Conjecture. According to the lemma above it is equivalent to all prime numbersof the form 4 q + 5 lie in N . N does not contain perfect squares. Here we prove that N does notinclude perfect squares, we need to use Jacobi’s symbol. Let p be an odd primeand n ∈ Z with ( n, p ) = 1, Legendre’s symbol is defined by (cid:18) np (cid:19) = (cid:26) , if n is a quadratic residue mod p , − , if n is a quadratic non-residue mod p .Let the standard factorization of m be p p · · · p k where the p r may be repeated.If ( n, m ) = 1 then Jacobi’s symbol is defined in term of the Legendre symbol by M. BELLO-HERN´ANDEZ, M. BENITO, AND E. FERN´ANDEZ (cid:0) nm (cid:1) = Q kj =1 (cid:16) np j (cid:17) . The following properties of the Jacobi symbol are very wellknown: Let m and m ′ be positive odd integers. (i.) If n ≡ n ′ (mod m ) and( n, m ) = 1, then (cid:0) nm (cid:1) = (cid:16) n ′ m (cid:17) . (ii.) If ( n, m ) = ( n, m ′ ) = 1, then (cid:0) nm (cid:1) (cid:0) nm ′ (cid:1) = (cid:0) nmm ′ (cid:1) . (iii.) If( n, m ) = ( n ′ , m ) = 1, then (cid:0) nm (cid:1) (cid:16) n ′ m (cid:17) = (cid:16) nn ′ m (cid:17) . (iv.) If n ≡ − /n ) = −
1. (v.)
Law of reciprocity.
Let m and n be odd coprime. Then (cid:0) nm (cid:1) (cid:0) mn (cid:1) = ( − n − m − . (vi.) If n, d are coprime positive integers and m satisfies n ≡ − m (mod d ), then (cid:0) dn (cid:1) = (cid:0) dm (cid:1) , see [10], p. 305. The Kronecker symbol isdefined in term of Jacobi’s symbol. It satisfies (cid:0) nm (cid:1) = 0 for ( n, m ) ≥ n , satisfying (4 abc − d = ( a + b ) n for some positive integers a, b, c, d with ( n, abd ) = 1, are no perfect square. Observethe symmetry of (4 abc − d = ( a + b ) n in a and b and Yamamoto proves, using ournotations, (cid:0) n ad (cid:1) = − abc − d = ( a + b ) n ,then (4 abc − d ′ = ( a + b ) n with d ′ = dn . Our class N contains numbers n whichare not included in Yamamoto class. For example, it is the case of n = 2009 , where a = 1, b = 293, c = 12, and d = 42, meanwhile α + 1 = 42 , β + 3 = 7 , and4 γ + 3 = 7. In fact, we have 2009 = 42 · · − − , · · − , (2009 , ·
42) = 7 . Lemma 7. N does not contain perfect squares.Proof. Let n ∈ N . Then there exist nonnegative integers α, β, γ such that n + (4 β + 3) = ( α + 1) ((4 β + 3)(4 γ + 3) − def = ( α + 1) τ. where τ = (4 β + 3)(4 γ + 3) −
1. Because of (4 β + 3 , τ ) = 1, we also have ( n, τ ) = 1and ( n + τ, τ ) = 1. Using reciprocity law for Jacobi’s symbol, we obtain(20) (cid:18) nn + τ (cid:19) (cid:18) n + τn (cid:19) = ( − n − n + τ − = 1 , here we have used n ≡ n + τ ≡ τ (mod n ), we have (cid:0) n + τn (cid:1) = (cid:0) τn (cid:1) . Taking into account n ≡ − (4 β + 3) (mod τ ) and property (vi) for the Jacobi’ssymbol cited above, we have (cid:0) τn (cid:1) = (cid:16) τ β +3 (cid:17) . Since τ ≡ − β + 3) and usingProperty (iv), we obtain (cid:16) τ β +3 (cid:17) = (cid:16) − β +3 (cid:17) = − . Therefore, from (20) we have (cid:16) nn + τ (cid:17) = − , it yields n is not a perfect square. (cid:3) Since 4( n + n −
1) + 5 = (2 n + 1) , and 4 q ( α, β, γ ) + 5 = p ( α, β, γ ), where p ( α, β, γ ) and q ( α, β, γ ) are given by (3) and (16) respectively, the Lemma 7 givesimmediately the following result: Corollary 2.
The numbers n + n + β +1 , ( n, β ) ∈ Z ≥ , have no divisors congruentto β + 2 modulo β + 3 . q − Strong Conjecture.
Lemma 7 and computer calculations let us to for-mulate the following conjecture which implies ESC and has been checked up to2 × . N EGYPTIAN FRACTIONS 9 q − Strong Conjecture:
The positive integers Z > is disjoint union of the threesets: • A = { q : ∃ n ∈ Z > , q = n + n − } . • B = { q : ∃ ( α, β, γ ) ∈ Z ≥ , q = q ( α, β, γ ) } , where q ( α, β, γ ) is given by (16). • C = { , , , , , , , , , , , , , , , , , , , , , , , , , , , , } . Of course this conjecture implies the q − Conjecture.5.
A Greedy-Type Algorithm for the Erd˝os-Straus Decomposition
Let us describe an algorithm to construct a decomposition of n as sum of atmost three Egyptian fractions. Using this algorithm we have checked ESC for n upto 2 × . In Lemma 8 bellow we prove that the following algorithm convergesfor all numbers n = abc − a − b with b ≡ c mod 4. Meantime in Lemma 9 we givesome conditions which characterize the convergence of this algorithm. A Greedy-Type Algorithm for the Erd˝os-Straus Decomposition: given n ∈ N , n ≥ , Step 1. Set j = 1 and q = ⌊ n ⌋ , the integer part of n .Step 2. Set x j = q + j , κ j = n − x j , and y j = ⌈ κ j ⌉ .Step 3. If n − x j − y j = 0 is true, the algorithm stops else we ask if z j = n − xj − yj ∈ Z is true, then the algorithm stops also and n = x j + y j + z j else we set j ← ֓ j + 1 and return to Step 2.Here y = ⌈ κ j ⌉ denotes the ceiling function evaluated at κ j .We call this algorithm “The Greedy-type Algorithm for Erd˝os-Straus decom-position.” Observe that the algorithm finds the greedy decomposition of j − nx j ,j = 1 , , . . . , and also stops when the decomposition is the sum of two Egyptianfractions. In fact, if r j ≡ − nx j (mod 4 j −
1) with r j ∈ [1 , j − j − nx j = 1 y j + r j nx j y j . The algorithm stops when r j divides nx j y j and z j = nx j y j r j . Lemma 8.
The greedy-type algorithm converges for all n = abc − a − b , with bc ≡ mod .Proof. If b ≡ c ≡ n = abc − a − b ≡ n hasa decomposition as sum of two Egyptian fraction. So we only have to consider b ≡ c ≡ x, z ) ∈ Z > , then 1 /x is the Egyptian fraction nearestto x + z and less than this number if and only if(21) x ( x − ≤ z. Moreover, we have(22) 4 n = 4 abc − a − b = 1 a bc − + 1 a ( ac − bc − + 1( ac − bc − n , and a bc − < a ( ac − bc − < ( ac − bc − p since c = 4 z + 3 ≥ ac − ≥ a , and n = abc − a − b ≥ a ⇔ b ( ac − ≥ a .According to (21) and (22), it is enough to prove a bc − (cid:18) a bc − − (cid:19) ≤ ( ac − bc − n ⇔ a (cid:18) a bc − − (cid:19) ≤ ( ac − abc − a − b ) . Observe that both relations a (cid:0) a bc − − (cid:1) = ba c/ − a / − a and ( ac − b − a ( ac −
1) are linear function in b , so comparing their leading coefficients and theirvalues at b = 1, the above inequality follows. (cid:3) Remark . We are checked that the algorithm given above also works for Sierpi´nski’sConjecture, n = x + y + z , taking q = ⌊ n ⌋ and making other few trivial changes,for n up to 10 .If n = 4 q + 1 and (4 q + 1)( q + j ) = ( s + 1)(4 j −
1) + r − (4 j − q + 1 − q + j − s + 1 = 4 j − − r (4 q + 1)( q + j )( s + 1)and(24) (4 q + 1) ( q + j ) = (4 q + 1)( q + j )( s + 1)(4 j −
1) + ( r − (4 j − q + 1)( q + j ) . So the first part of following result follows:
Lemma 9.
Set n = 4 q + 1 ∈ Z > . Let s = s ( q, j ) and r = r ( q, j ) be such that (4 q + 1)( q + j ) = s (4 j −
1) + r , ≤ r ≤ j − for some j ∈ Z > . Then thegreedy-type algorithm converges for n = 4 q + 1 if and only if one of the followingequivalent statement holds: (1) (4 j − − r divides to (4 q + 1)( q + j )( s + 1) . (2) (4 j − − r divides to (4 q + 1) ( q + j ) .Proof. According to (23), it is enough to check that if (4 j − − r divides to(4 q + 1) ( q + j ) , then (4 j − − r also divides to (4 q + 1)( q + j )( s + 1). Let δ = (4 j − , s + 1), so we have ( α, β ) = 1, where α = j − δ and β = s +1 δ . Since δ isodd, (4 q + 1)( q + j ) = (4 q + 1) q +1+(4 j − , and(25)(4 q +1)( q + j )( s +1) = ( s +1) (4 j − − (4 k − − r )( s +1) = ( s +1) δα − ( s +1) δ ( α − β ) , if δ divides (4 q + 1)( q + j )( s + 1), then δ divides (4 q + 1)( q + j )( s + 1). On the otherhand, using (24) and the hypothesis we obtain α − β divides (4 q + 1)( q + j )( s + 1).Therefore, from (25) we conclude that (4 j − − r = δ ( α − β ) divides (4 q + 1)( q + j )( s + 1), and the lemma is proved. (cid:3) N EGYPTIAN FRACTIONS 11
Remark . For theoretical considerations is useful to observe that according to ournotations (4 q +1)( q + j ) = s (4 j − r ⇔ (4 q +1) +(4 q +1)(4 j −
1) = 4 s (4 j − r ,so we have (4 q + 1) = (4 s − (4 q + 1))(4 j −
1) + 4 r def = (4 k − j −
1) + 4 r and(4 q + 1) ≡ r mod 4 j − Remark . From statement 2 in the lemma above it follows that if n = 4 q + 1 isa prime, the greedy-type algorithm stops at a step j with 4 j − < q + 1 if andonly if 4 j − − r divides ( q + j ) . Remark . Given any j ∈ Z > there exists a prime number n = n j such that thenumber of steps in the algorithm above is larger than j . In fact, if n def = 4 lcm { , . . . , j − , , , . . . , j − } t + 1 def = 4 q j + 1is a prime for a given t ∈ Z > , then for each k ≤ j , we have(4 q j + 1)( q j + k ) ≡ k, mod 4 k − , and since ( k, k −
1) = 1 , k − − k = 3 k − ∤ (4 q j + 1) ( q j + k ) . Therefore, according to the Lemma 9, the greedy-type algorithm does not convergein the first j steps for n j . Appendix. Computer Programs
In this section we include some programs which are used to check the Erd˝os-Straus Conjecture, q − Conjecture and the q − Strong Conjetures. For an indepen-dent reading we set again some facts already cited.6.1.
A computer program checking the q − Strong Conjecture.
Lemma 7and computer calculations let us to formulate the following conjecture which impliesthe ESC and has been checked up to 2 × : q − Strong Conjecture:
The set of positive integers, Z > , is disjoint union ofthe following three sets: • A = { q : ∃ n ∈ Z > , q = n + n − } . • B = { q : ∃ ( α, β, γ ) ∈ Z ≥ , q = q ( α, β, γ ) } , where(26) q = q ( α, β, γ ) = ((4 β + 3) γ + (3 β + 2))( α + 1) − ( β + 2) , • C = { , , , , , , , , , , , , , , , , , , , , , , , , , , , , } . Of course this conjecture implies the q − Conjecture, see below and Lemma 6. Nextwe describe our algorithm:
Step 1:
We generate a matrix w of equivalence classes in the set B using thefunction cribata of two arguments. The first column of w are the moduliand the second are their corresponding rests. We only use moduli which aredivisors of a given number. The values in q of two of the three parameters α, β, and γ , or their associated new parameters after a change of variable, are setting up to a bound. We use several equivalent expressions to thefunction (26):(27) q = 4 xyz + 3 xy + 3 xz + 2 x + 4 yz + 2 y + 3 z,q = x (4 yz + 3 y + 3 z + 2) + 4 yz + 2 y + 3 z,q = y (4 xz + 3 x + 4 z + 2) + 3 xz + 2 x + 3 z,q = z (4 xy + 3 x + 4 y + 3) + 3 xy + 2 x + 2 y, q + 6 + x = (4 y + 3)(4( x + 1) z + 3 x + 2) ,q + y + 2 = ( x + 1)((4 y + 3) z + 3 y + 2) , (3 + 4 z ) q + 4 + 5 z = ((3 + 4 z ) x + 4 z + 2)((3 + 4 z ) y + 3 z + 2) , and others which are easy to find after a change of variables, setting y = z + a in (27), we have q = x (4 z + z (4 a + 6) + 3 a + 2) + 4 z + z (4 a + 5) + 2 a, doing z = u − a − in (27) with u , a of different parity, − u < a < u , q = u + u − − a − a + u + 12 + x ( u + u − a ) , setting a = u − d − q = ( u − d − d + 2) + 3 d + x ( u (4 d + 3) − (2 d + 1) ) , ≤ d ≤ u − , for a = − u + 2 d + 1, we have q = ( u − d − d + 3) + 2 d + x ( u (4 d + 3) − (2 d + 1) ) , ≤ d ≤ u − , doing x = s − t , y = t − z in (27), we obtain q = s ( − z + 4 tz + 3 t + 2) + 4 tz − z − t z + 4 tz + z − t , doing the same x = s − t , z = t − y in (27), q = ( − y + 4 ty + 3 t + 2) s + 4 ty − y − t y + 4 ty − y − t + t, with y = s − t , z = t − x in (27), q = ( − x + 4 tx − x + 4 t + 2) s + 4 tx − x − t x + 4 tx − x − t + t, doing y = s − t , x = t − z in (27), q = s ( − z + 4 tz + z + 3 t + 2) + 4 tz − z − t z + 2 tz + z − t , setting z = s − t , y = t − x in (27), q = s ( − x + 4 tx − x + 4 t + 3 + 4 tx − x − t x + 4 tx − t − t, with z = s − t , x = t − y in (27), q = s ( − y + 4 ty + y + 3 t + 3) + 4 ty − y − t y + 2 ty − t − t. Step 2:
Next, we sieve q in the residue classes in the vector w using thefunction cricu .The steps 1 and 2 are done two times. First, we use moduli which aredivisors of the product of all primes up to 19. Next, we add the factor 23obtaining tc = 2 × × × × × × × × . N EGYPTIAN FRACTIONS 13
Step 3:
As a by-product of steps above we have many rests modulo tc whichare not sieved, all these are saved in the vector v . The function adjunta attaches to a matrix a vector, this is also used in the Step 1 generatingthe matrix w . At the last, we use the function fasf which combines squarechecking with functions cri throughout the equivalent classes in v modulo tc .Next, we show Sage code of our program, although we write also our algorithmin UBASIC and Pari GP. The Sage code is more transparent and easy to read.Moreover, the program have been run on the operating systems: MS-DOS, Linux,and OS X. def adjunta(s,r,u):l=len(u)h=0for i in xsrange(l):if s%u[i][0]==0:c=gcd(s,u[i][0])if r%c==u[i][1]:h=1if h==0:l=l+1u.append([s,r])return(u)def cribata(t,m):u=[[2,0],[3,0],[7,0]]m1=m+1for b in xsrange(m1):for a in xsrange(b+1):s=4*a*b+3*a+3*b+2if t%s==0:r=4*a*b+2*a+3*bu=adjunta(s,r,u)s=4*a*b+3*a+3*b+2if t%s==0:r=4*a*b+3*a+2*bu=adjunta(s,r,u)s=4*a*b+3*a+4*b+2if t%s==0:r=3*a*b+2*a+3*bu=adjunta(s,r,u)s=4*a*b+4*a+3*b+2if t%s==0:r=3*a*b+3*a+2*bu=adjunta(s,r,u)s=4*a*b+3*a+4*b+3if t%s==0:r=3*a*b+2*a+2*b u=adjunta(s,r,u)s=4*a*b+4*a+3*b+3if t%s==0:r=3*a*b+2*a+2*bu=adjunta(s,r,u)s=4*a*b-4*a^2+3*b+2if t%s==0:r=4*a^2*b-4*a^2-4*b^2*a+4*a*b-a-3*b^2+bu=adjunta(s,r,u)s=4*a*b-4*a^2+3*b+2if t%s==0:r=4*a^2*b-4*a^2-4*b^2*a+4*a*b+a-3*b^2u=adjunta(s,r,u)s=4*a*b-4*a^2+4*b+2-aif t%s==0:r=4*a^2*b-3*a^2-4*b^2*a+4*a*b-a-4*b^2+bu=adjunta(s,r,u)s=4*a*b-4*a^2+a+3*b+3if t%s==0:r=4*a^2*b-3*a^2-4*b^2*a+2*a*b-3*b^2-bif r>0:u=adjunta(s,r,u)s=4*a*b-4*a^2+a+3*b+2if t%s==0:r=4*a^2*b-3*a^2-4*b^2*a+2*a*b-3*b^2+aif r>0:u=adjunta(s,r,u)s=4*a*b-4*a^2-a+4*b+3if t%s==0:r=4*a^2*b-3*a^2-4*b^2*a+4*a*b-4*b^2-bif r>0:u=adjunta(s,r,u)s=-4*a^2+4*a*b-4*a+3*b-1if s>0:if t%s==0:r=- 4*a^2 + 4*a*b - 5*a + 3*b - 3if r>0:u=adjunta(s,r,u)s=-4*a^2+4*a*b-4*a+3*b-1if s>0:if t%s==0:r= - 4*a^2 + 4*a*b - 3*a + 2*b - 2if r>0:u=adjunta(s,r,u)return(u)t32=1; N EGYPTIAN FRACTIONS 15 t43=1;for p in primes(2,500000):if p%3==2:t32=t32*pif p%4==3:t43=t43*pdef cri(q):j=-1h=0while h==0 and j N EGYPTIAN FRACTIONS 17 a0=-1for l1 in xsrange(0,v2):for i in xsrange(0,v1):q=v7[i]+l1*tcricu(q)v1=len(v)print v1save(v,"qf23")a0=0for l1 in xsrange(0,9000):for i in xsrange(0,v1):q1=v[i]+l1*tcwt=fasf(floor(q1))a=wt[1]if wt[0]==0:print q1," falla "a=0if a>a0:a0=aprint q1,a A computer program verifying the q − Conjecture and Erd˝os-StrausConjecture of Type-I. According to Lemma 6 if for each q ∈ Z > there exist x, y, z in Z ≥ such that one of the following relations holds q = 1 + 3 x + 3 y + 4 xy,q = 5 + 5 x + 5 y + 4 xy,q = ((4 y + 3) z + (3 y + 2))( x + 1) − ( y + 2) . the Erd˝os-Straus Conjecture is true. We conjecture that all the natural number q ∈ Z > can be written as one of the three above relations. We refer it as the q − Conjecture . Observe that this is equivalent to for all prime numbers of the form4 q + 5 the corresponding value q can be write as q = ((4 y + 3) z + (3 y + 2))( x + 1) − ( y + 2) for some x, y, z in Z ≥ . It is worst to observe that the identity4 abc − a − b = 1 a bc − + 1 a ( ac − bc − + 1( ac − bc − n , with n = abc − a − b, implies that n has a decomposition as sum of three Egyptianfraction with exactly one of the denominators multiplies of n (a decomposition oftype I). We are verified q − Conjecture for q ≤ . × . For this object, we doa litter change in the program above. We write two new functions crice and fase instead of cricu and fasf , respectively.The Sage code of these function is: def crice(q):if gcd(4*q+5,tc)==1:if gcd(q+2,t3)==1:if gcd(3*q+4,t3)==1:if gcd(2*q+3,t4)==1: if gcd(q+3,t7)==1:if gcd(7*q+9,t7)==1:if (q+4)%19>0:if (11*q+14)%19>0:if (q+5)%11>0:if (15*q+19)%11>0:if (q+7)%17>0:if (23*q+29)%17>0:h=0j=-1while h==0 and j We areverified that all the natural numbers n , 2 ≤ n ≤ × , satisfy the Ed˝os-StrausConjecture. To this end we do a more sharp sieve than in the above programs. Thenew cricue function replace crice function and add several new testing taking intoaccount Lemma 9 which gives a characterization for the convergence of the greedy-type algorithm. Some of these new checkup are given by the minimal number ofsteps needed for the convergence of the greedy-type algorithm. Moreover, we divide N EGYPTIAN FRACTIONS 19 the interval [2 . × , × ] in several intervals each one of 10 numbers, thenusing the function ferti , which uses fase function, we check the ESC in each interval.Next we show the Pari GP code of our program: fase(q,h,q0,a1,a0,p,q7)={h=0;if(gcd(4*q+5,tp)>1,h=1,if(gcd(2*q+3,t43)>1,h=1,if(gcd(q+2,t32)>1,h=1,if(gcd(3*q+4,t32)>1,h=1,if(cri(q)==1,h=1;a0=2,if(isprime(4*q+5)==0,h=1;a0=2,q7=(3*q)\2;q0=q+1;a1=-1;a0=-1;h=0;while(h==0&&a0 N EGYPTIAN FRACTIONS 23 h=1,))),);if(h==0,j23++;v23[j23]=q,)));x=0;v=vector(j23,x);for(i23=1,j23,v[i23]=v23[i23]);v1=matsize(v)[2];print("erti23 ",v1); write("v.txt",v)}{tp=1;t32=1; t43=1;forprime(p=2,15000,tp*=p;if(p%3==2,t32*=p,);if(p%4==3,t43*=p,))}{t=2*27*5*7*11*13*17*19*23;w=cribata(t,100);w1=matsize(w)[1];print (w1)}ferti(li,lf,a0,l0,l,i7,q,y,a)={a0=0;l0=matsize(v)[2];for(l=li,lf,print(l);for(i7=1,l0,q=v[i7]+t*l;y=fase(q);a=y[2];if(y[1]==0,print(q," falla"); write("salida23",q,"falla"),if(a>a0,a0=a; print(" ",q," ",a);write ("salida23"," ",q," ",a),))))} Acknowledgment We wish to thank J´onathan Heras Vicente, Jos´e Antonio Mart´ınez Mu˜noz andJuan Luis Varona, who allowed us to use their computers to perform calculations. 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Classical theory. Cambridgetracts in advanced mathematics. 97. (2007).[13] Mordell, L. J. Diophantine Equations. London: Academic Press, 1969.[14] Niven, I., Zuckerman, H.S., Montgomery, H.L. An Introduction to Theory of Numbers. FithEd. John Wiley & Sons. 1991.[15] Schinzel, A. On sums of three unit fractions with polynomial denominators. Funct. Approx.Comment. Math. 28, 2000, 187-194. [16] Sierpi´nski, W. Elementary Theory of Numbers. Second ed. revised and enlarged by A.Schinzel, coedition between Elsevier Science Publishers B.V. and PWN-Polish Science Pub-lishers, 1988.[17] Swett, A. The Erd˝os-Strauss Conjecture. web: http://math.uindy.edu/swett/esc.htm. Rev.10/28/99[18] Vaughan, R.C. On a Problem of Erd˝os, Straus, and Schinzel. Mathematika 17, 1970, 193–198.[19] Yamamoto, K. On the Diofantine Equation n = x + y + z . Mem Fac. Sci. Kyushu Univ.Ser. A, V. 19, No. 1, 1965, 37–47. DPTO. DE MATEM ´ATICAS Y COMPUTACI ´ON, UNIVERSIDAD DE LA RIOJA, EDIF.J. L. VIVES, CALLE LUIS DE ULLOA S/N, 26004 LOGRO ˜NO, SPAIN E-mail address : [email protected] INSTITUTO SAGASTA, GLORIETA DEL DOCTOR ZUB´IA, S/N, 26003, LOGRO ˜NO,LA RIOJA, SPAIN E-mail address : [email protected] INSTITUTO SAGASTA, GLORIETA DEL DOCTOR ZUB´IA, S/N, 26003, LOGRO ˜NO,LA RIOJA, SPAIN E-mail address ::