aa r X i v : . [ m a t h . G M ] S e p On the numbers that are sums of three cubes
N.D. BagisAristotele University of Thessaloniki-AUTHThessaloniki-Greecee-mail:[email protected]
Abstract
We examine what integers are representable as sums of three cubes.We also provide formulas for the number of representations of x + y + z = n under the condition x + y + z = t . Also we show howthe problem of three cubes is related to abc − conjecture. keywords : Sums of three cubes; Diophantine equations; Higherforms; Representations of integers We will proceed to find all integer solutions ( x, y, z ) of x + y + z = n (1)with height x + y + z = t. (2)The above problem (without condition (2)) is rather difficult and is open formore than hundred years years. In recent days, with the help of computersmany unknown results come to light. For example it simple to state that1 + 1 + 1 = 4 + 4 + ( − = 3 . But it is very difficult to find that569936821221962380720 + ( − ++( − = 3 . In 1992, Roger Heath-Brown conjectured that every n unequal to 4 or 5 modulo9 has infinitely many representations as sums of three cubes. This conjectureuntil now days remains unproved. 1 Analyzing the problem.
Set x + y + z = n (3)The cases (1) together with (2) are solvable with Wolfram alpha.Set x + y + z = t , xy + yz + zx = r , xyz = s. (4)Then x + y + z = t − r = n . (5)Hence t = ±√ n + 2 r (6)Also then x + y + z = n ⇔ s − rt + t = n. (7)Setting the value of (6) to (7) we get n ± ( r − n ) √ r + n = 3 s. (8)We set also p be such that p = 3 s − n. (9)Equation (7) becomes finally t ( t − r ) = − p, (10)which is solvable (however this means nothing since for solving (10) we need toknow where p exactly varies. Clearly we have | xy + yz + zx | ≤ x + y + z ⇔ | r | ≤ n (11)Also hold the following identity of Euler x + y + z − xyz = ( x + y + z )( x + y + z − xy − yz − zx ) (12)Hence n − s = t ( n − r ) ⇔ − p = t ( n − r ) ⇔ pt + n = r ⇒ (cid:12)(cid:12)(cid:12) pt + n (cid:12)(cid:12)(cid:12) = | r | ≤ n Hence − n ≤ pt ≤ t = n + 2 r ⇔ t − n r ⇒ (cid:12)(cid:12)(cid:12)(cid:12) t − n (cid:12)(cid:12)(cid:12)(cid:12) = | r | ≤ n ⇒ − n ≤ t − n ≤ n (14)2ence 0 ≤ t ≤ n and r ≥ − n ⇒ pt ≥ − n − n ⇒ ≥ pt ≥ − n . Alsofrom (7) n − s = t ( n − r )Hence given x + y + z = n, with height x + y + z = t If we set ǫ = n − r = | n − r | ≥ , then from (7) we get n − s = tǫ ⇔ ǫ = t − r and we have to solve the system3 tr − s = t − n , n + 2 r = t . (15)with respect to r, s, n . We have solution if gcd(3 t,
3) = 3 | ( t − n ) r = C , s = tC + n − t n = − C + t , C ∈ Z . (16)Now consider the product( x − t )( y − t )( z − t ) = xyz − t ( xy + yz + zx ) + t ( x + y + z ) − t == xyz − t ( xy + yz + zx ) = s − tr == tC + n − t − tC = n − t x − t )( y − t )( z − t ) = n − t , (17)we are done. Set x = A + t , y = B + t , z = C + t and the equation (17) becomes ABC = n − t . (18)The general equation ABC = N have L ( N ) = 4 P The equation x + y + z = n , with height x + y + z = t , n = t , have finitenumber of solutions. Moreover the problem (1),(2) is equivalent to ABC = n − t A + B + C = − t. (20)Actually A = x − t , B = y − t , C = z − t , t = x + y + z .Hence given n, t , equation (18) have L ( N ) solutions, where L ( N ) := 4 X d | N τ ( d ) (21)and N = (cid:12)(cid:12)(cid:12) n − t (cid:12)(cid:12)(cid:12) . Taking one by one the L ( N ) solutions of (18), we evaluateall cases such A + B + C = − t and solve (1),(2) according to Theorem 1.Also we have the next Theorem 2. Given integers n, t , with n = t , the number of representations of the integer n as x + y + z = n , ( x, y, z ) ∈ Z , (22)with height x + y + z = t, (23)is R ( t, n ) = X = abs( d ) | abs( N ( t, n ))0 = abs( δ ) | abs( d ) δ + d/δ + N ( t, n ) /d = − t , (24)where N ( t, n ) = n − t . Note that if n − t = 0(mod3), we don’t have represen-tations. Also if n = t , we have infinite representations. In this case (and forconvenience) we set in 0 in (24). Theorem 3. Assume that n is a fixed integer, then the number of representations of n in theform x + y + z = n , ( x, y, z ) ∈ Z , (25)4ith − j ≤ x + y + z ≤ j , where j ≥ 0, is C ( j, n ) = j X t = − j R ( t, n ) , (26)where R ( t, n ) is that of (24). Moreover if n is an integer such that n − t = 0and n − t ≡ n in (25)is lim j → + ∞ C ( j, n ) . Note. Assume that exists n , t ∈ N such that R ( t, n ) = 0, for all t with | t | ≥ t > t < t , we have R ( t , n ) ≥ i) If n ≡ t ≡ , R ( t , n ) = 0, which is not true.Hence t ≡ ii) If n ≡ t ≡ , R ( t , n ) = 0, which is nottrue. Hence t ≡ iii) If n ≡ t ≡ t ≡ n (mod3) . (27)If N = n − t and n − t ≡ d runs through allinteger divisors of N and δ is divisor of d and hence of N , in order to havesolution, we must have δ + d/δ + N /d = − t. (28)Solving with respect to δ the above equation, we get δ = − N d + t ± s − d + (cid:18) N d + t (cid:19) . (29)Hence N must be even. However exist cases such N is even and we have nosolution i.e. t = − n = 4, N = 4. Hence when N is odd we have nosolutions. In general relation (29) give us the following Theorem 4. Given n , t integers, we set N := n − t . Then the number of representations inthe form x + y + z = n, (30)with height x + y + z = t, (31)is R ( t, n ) = − X < d | N ( N /d + 2 t ) = 4 d X =abs( d ) | N S Z + (cid:18) N d + 2 t (cid:19) − d ! , (32)5here S Z + ( n ) is 1 if n is a nonnegative square of integer and 0 otherwise. Proof. In order to have a representation added in the sum (24), a divisor d of N = n − t must be such x d ± q x d − d ∈ Z , (33)where x d = N d + t . Hence the number of representations becomes R ( t, n ) = X = abs( d ) | N x d ± p x d − d ∈ Z − X = abs( d ) | N x d = d . (34)Here we take the negative part to avoid zero discriminants. Hence if u = x d − p x d − d and v = x d + p x d − d , then uv = d and u + v = 2 x d and we also have R ( t, n ) = 3 X = abs( d ) | N < | u | ≤ d < | v | ≤ duv = du + v = N /d + 2 t . (35)However the equivalence (28) ⇔ (29) is that which reduces the problem. Lemma 1.1) If n = 9 k , then i) If k = 4 k in order to have solutions of (1), we must have t = 6 l . ii) If k = 4 k + 1 then t = 6 l + 3. iii) If k = 4 k + 2, then t = 6 l . iv) If k = 4 k + 3, then t = 6 l + 3. If n = 9 k + 1, then i) If k = 4 k , then t = 6 l + 1. ii) If k = 4 k + 1, then t = 6 l + 4. iii) If k = 4 k + 2, then t = 6 l + 1. iv) If k = 4 k + 3, then t = 6 l + 4. If n = 9 k + 2, then i) If k = 4 k , then t = 6 l + 2. ii) If k = 4 k + 1, then t = 6 l + 5. iii) If k = 4 k + 2, then t = 6 l + 2. iv) If k = 4 k + 3, then t = 6 l + 5. If n = 9 k + 3, then i) If k = 4 k , then t = 6 l + 3. ii) If k = 4 k + 1, then t = 6 l . iii) If k = 4 k + 2, then t = 6 l + 3. iv) If k = 4 k + 3, then t = 6 l . If n = 9 k + 6, then i) If k = 4 k , then t = 6 l . 6 i) If k = 4 k + 1, then t = 6 l + 3. iii) If k = 4 k + 2, then t = 6 l . iv) If k = 4 k + 3, then t = 6 l + 3. If n = 9 k + 7, then i) If k = 4 k , then t = 6 l + 1. ii) If k = 4 k + 1, then t = 6 l + 4. iii) If k = 4 k + 2, then t = 6 l + 1. iv) If k = 4 k + 3, then t = 6 l + 4. If n = 9 k + 8, then i) If k = 4 k , then t = 6 l + 2. ii) If k = 4 k + 1, then t = 6 l + 5. iii) If k = 4 k + 2, then t = 6 l + 2. iv) If k = 4 k + 3, then t = 6 l + 5. Remarks. Actually it is known that when n = 9 k ± 4, then equation (1)have no representations at all. Proof of Lemma 1.1) For n = 9 k we have i) Assuming n = 9 k , k = 4 k , then using 3 | ( n − t ) and N = n − t = even weget easily the first result. ii) For n = 9 k , k = 4 k + 1, then using 3 | ( n − t ) and N = n − t = even we getthe second result. iii) The third result ( n = 9 k , k = 4 k + 2) follows in the same way . . . etc. Theorem 5. When n ≡ ± Proof. From Theorem 4 relation (32) we have (in order to have solutions) (cid:18) N d + 2 t (cid:19) − d = k , ( a )where d is any divisor of N . From relation ( a ) we have that exists integer y such that 4 d = y − k . ( b )Clearly k, y are both even or both odd. Assume that both k, y are odd, then k = 2 v + 1, y = 2 u + 1. Then 4 d =(4 u + 4 u + 1) − (4 v + 4 v + 1) = 4( u + u − ( v + v )). Hence d = u + u − v − v. Relation ( a ) can be written in the form N = yd − td ⇔ n − t = 3 yd − td ⇔ = 3(2 u + 1)( u + u − v − v ) − t ( u + u − v − v ) + t . ( c )From Lemma 1 we have if n = 9 s + 4, s even, then t = 4 + 6 l . Hence n =4 + 18 s ≡ t ≡ t ≡ c ) becomes n ≡ u + 1)( u + u − v − v ) − u + u − v − v ) + 10(mod18) . As one can see by taking all the possible cases u ≡ , , . . . , v ≡ , , . . . , n ≡ , n is not 4(mod18).In case y = 2 u , k = 2 v , then d = u − v and n = 3( u − v )(2 u ) − t ( u − v ) + t . ( d )This case lead us again to n ≡ , n ≡ ii) From Lemma 1, the case n = 9 s +4, s odd is n ≡ t ≡ t ≡ t ≡ y = 2 u + 1, k = 2 v + 1) n ≡ u + u − v − v )(2 u + 1) − u + u − v − v ) + 1(mod18) . Taking all the possible cases we see that n ≡ , n ≡ y = 2 u , k = 2 v , we have n = 3( u − v )(2 u ) − u − v ) + 1(mod18) ≡ , . Hence again is not proper. For the case n = − s , s even, we have t = 5+6 l . Hence n ≡ − ≡ t ≡ t ≡ t ≡ y = 2 u + 1, k = 2 v + 1, then n ≡ u + u − v − v )(2 u + 1) − u + u − v − v ) + 17(mod18)and this leads to n ≡ , y = 2 u , k = 2 v , we have n ≡ u − v )(2 u ) − u − v ) + 17(mod18) ≡ , ii) The case n = − s , with s odd give us n ≡ t ≡ t ≡ t ≡ Theorem 6. If N = n − t and exists u, v integers such that N = (2 u + 1)( u + u − v − v ) − t ( u + u − v − v ) (36)or N = 2 u ( u − v ) − t ( u − v ) (37)then we have a solution of (1),(2) (+1 or +2 added in R ( t, n )). In case thatboth (36),(37) are not satisfied then R ( t, n ) = 0.i) In case of (37), with k = 0 then we have at least one solution (+1 in R ( t, n )).ii) If k = 0, then for each ( u, v ) satisfying (37) we add +2 in R ( t, n ).iii) For each ( u, v ) satisfying (36) we have +2 in R ( t, n ).iv) In any case in order to have +1 or +2 in R ( t, n ), N must be even and if d | N , then d must be of the form 4 d = y − k , with | y | = | N /d + 2 t | . Corollary 1. R (0 , n ) > 0, whenever n ≡ d of n/ n d − d = perfect square . (38)Assume now the equation 9 d ( k + 4 d ) = n (39)where n, k, d are in general integers. Assume L ( n ) denotes the number ofsolutions of (39) for a given n . If we where able to prove that exists se-quence of natural numbers n m , m = 1 , , . . . , such that n m were increasingwith lim m → + ∞ L ( n m ) = + ∞ i.e. the number of solutions of (39) for n = n m will be growing without a bound as m goes to infinity. Then in view of Theorems2,3,4, the infinite numbers n = n m will correspond to x + y + z = n m , x + y + z = 0, m = 1 , , . . . (40)and have L ( n m ) = R (0 , n m ) number of solutions andlim m →∞ R (0 , n m ) = + ∞ . (41)Hence the equations x + y + z = n m , m = 1 , , . . . , (42)9ith L ∗ ( n m ) being the number of solutions of the m − th equation, will have atleast finite number of solutions (from Theorem 3). But from (41) and Theorem3 we will have lim m → + ∞ L ∗ ( n m ) = + ∞ . (43)Hence, as m → + ∞ , the number of solutions of (42) will then goes to infinity.Numerically we can construct such numbers n m . We can see also (numerically)that they satisfy the approximate relation σ ( n m ) ≈ e γ n m log log n m . (44)Hence if we define the set S ( a, b ) = { n ∈ N : 5040 < a ≤ n ≤ b and 0 . < σ ( n ) e γ n log log n < } . (45)Then n = n m ∈ ( a, b ) iff n ∈ S ( a, b ), L ( n m +1 ) − L ( n m ) = 6 and n is the smallestpossible i.e. from all n ∈ N such that L ( n ) = c = const > c = 24, wechoose the smallest one n = 19440. We have calculated the first few valuesof n m . These are n = 90, L ( n ) = 12; n = 720, L ( n ) = 18 n = 19440, L ( n ) = 24; n = 55440, L ( n ) = 30; n = 443520, L ( n ) = 36; . . . . Note thatRiemann’s hypothesis is equivalent with the Robin − Ramanujan inequality σ ( n ) < e γ n log log n , ∀ n > , (46)where σ ν ( n ) = P d | n d ν is the sum of the ν − th power of positive divisors of n .But of course this is no proof that such n m are infinite. In the next section weprovide a proof based in abc − conjecture see Theorem 9 below.For to solve (39), we have n = 3 · n , where n = 2 a a . . . p a s s is the primedecomposition of n . Then d = 2 b b . . . p b s s , with c i = a i − b i ≥ i = 1 , , . . . , s is any divisor of n and (39) becomes l − k = 4 d , ∀ d | n . (47)This equation have classically 2 τ ( d ) number of solutions. But for these solutionswe must have the extra condition: l = 2 c c . . . p c s s or equivalently l = n/ (3 d ).In case t = t = 0, the equation x + y + z = n , x + y + z = t , is equivalent to (cid:18) n − t d + 2 t (cid:19) − d = k , where d is divisor of N = n − t . This becomes l − k = 4 d , ( l − t ) d = N . (48)10 The reduced problem Assume the equation xy ( x − y − 1) = n, (49)where x, y, n ∈ Z . I rewrite equation (49) in the form( − x ) y ( − x − y − 1) = n ⇔ xy ( − x − y − 1) = − n. Hence setting z = − x − y − 1, we get that (49) is equivalent to xyz = − n , x + y + z = − . (50)Now given n integer, according to Theorem 2, equation (49) and hence (50)have L ( n ) = R (cid:18) , − n (cid:19) = X = abs( d ) | n = abs( δ ) | abs( d ) δ + d/δ − n/d = − , (51)number of solutions. I will show that for certain choices of n we have always so-lution of (37). Set n = pp (1 + p + p ), then exists always a divisor d = pp > n and a divisor δ = p of d , such that δ + d /δ − n/d = − 1. Taking n = n ν = p ν p ν +1 (1 + p ν + p ν +1 ), where p ν is the ν − th prime, we have aninfinite choice of positive integers n = n ν such that L ( n ν ) > 0. Hence equation(49) ⇔ (50) have infinite number of n with at least one solution.Also in view of Theorem 4 and relation (34) we get the next Theorem 6. The equation (49) ⇔ (50) have number of solutions: L ( n ) = − X < d | n ( n/d − = 4 d X =abs( d ) | n S Z + (cid:18)(cid:16) nd − (cid:17) − d (cid:19) (52) Proof. The equation (49) ⇔ (50) have L ( n ) = X = abs( d ) | ny d ± p y d − d ∈ Z y d > d − X = abs( d ) | ny d = d, y d ∈ Z , (53)where y d = n d − number of solutions. Hence if u = y d − p y d − d and v = y d + p y d − d , then uv = d and u + v = n/d − 1. Also then L ( n ) = 3 X = abs( d ) | n < | u | ≤ d < | v | ≤ duv = du + v = n/d − . (54)11ssume now the form x y + xy = n , ( x, y ) ∈ Z , (55)with n ∈ Z . This is equivalent to xyz = n , x + y − z = 0, ( x, y, z ) ∈ Z (56)and this to xyz = − n , x + y + z = 0, ( x, y, z ) ∈ Z . (57)If we manage to show that (57) have infinite solutions ( x, y, z ) ∈ Z for aninfinite sequence n = n m ∈ N , m = 1 , , . . . and the number of solutions of each n = n m grows unboundedly. Then according to Theorem 3, equation x + y + z = n , will have also unboudeddly set of solutions. However (55) ⇔ (56) ⇔ (57).Assuming equation (55) with x, y > 0, then according to abc − conjecture, wehave that: for every ǫ > 0, exists a constant K ǫ > ǫ only suchthat (rad( n )) ǫ > K ǫ ( x + y ) , (58)where n := rad( n ) = Y p | np − prime p. (59)Assuming Sol (+) ( n ) = { ( x , y , z ) , ( x , y , z ) , . . . , ( x ν , y ν , z ν ) } is the set of pos-itive solutions of (57) and L (+) ( n ) = ν > L (+) ( n ) (rad( n )) ǫ > K ǫ ν X k =1 z k . Hence we have the next Theorem 7. Equation (57) ⇔ (56) ⇔ (55) have for each n positive integer ν = L ( n ) = R (0 , n ) = − X < d | nn = 4 d X =abs( d ) | n S Z + (cid:18) n d − d (cid:19) (60)solutions. Moreover in view of notes above we have1 ν ν X k =1 z k < K ǫ n ǫ , ∀ ǫ > . (61) Theorem 8. The equation x + y + z = − n , x + y + z = 0, n ≥ L ( n ) solutions ( x, y, z ) ∈ Z . Under the abc − conjecture, for a giveninteger n > n ≥ n have | x | + | y | + | z | < C ǫ n ǫ , (63)where ǫ > C ǫ positive constant depending on the choice of n and ǫ . Proof. Immediate consequence of Theorem 7, relation (58), the symmetry of (57) andTheorem 1.Continuing our arguments, we use the bounds conjectured in [1]: Conjecture 1. (Generalized abc − conjecture)If x, y, z are positive integers and k = rad( n ), where x + y = z , n = xyz , thenthere is a constant C such that z < k exp " s k log log k (cid:18) k k + C log log k (cid:19) (64)holds. Also there is a constant C such that z > k exp " s k log log k (cid:18) k k + C log log k (cid:19) (65)holds infinetly often.From the above notes together with Theorems 1,3 we can show that Theorem 9. Assuming the generalized abc conjecture, the equation x + y + z = n , ( x, y, z ) ∈ Z , (66)with n = 3 b m , have unbounded set of solutions as m → + ∞ . The numbers b m are defined as follows: If n = xy ( x + y ), z = x + y , k = rad( xyz ) and the triples( x, y, z ) ∈ Z ∗ satisfy (65), then b m = n = xyz . Proof. From Conjecture 1, since (65) is satisfied infinetly often, there exists certainsequence b m ∈ N such that n = b m = xyz , m = 1 , , . . . , z = x + y of integerswith lim b m = + ∞ . Also lim rad( b m ) = ∞ . That is, from Cauchy inequality z = x + y ≥ √ xy ⇒ z ≥ b m . Using (64) we get rad( b m ) → + ∞ . Alsolim L ( b m ) = ∞ i.e. (55) and hence (56) ⇔ (57) (with n = n = b m ) have un-bounded set of solutions. This can be proved as follows: Assume the equation1366), with n = b m ∈ Z ∗ + . We denote with Sol (+) m = { ( x , y , z ) , ( x , y , z ) , . . . , ( x ν , y ν , z ν ) } , the set of positive integer triples ( x, y, z ) = ( x, y, x + y ) who satisfy (56) when n = b m . Then assume that as m → + ∞ we have ν = ν m < ∞ (is bounded).But k = rad( xyz ) = rad( b m ) → + ∞ as m → ∞ . Also according to Conjecture1 we can write1 ν ν X i =1 z i > k exp " s k log log k (cid:18) k k + C log log k (cid:19) → + ∞ . If ν = ν m were bounded then 1 ν ν X i =1 z i < ∞ , which is impossible. Hence lim m →∞ ν m = + ∞ .Hence as consequence of Theorems 1,2,3, we have for n = 3 b m , that, the numberof solutions of (66) islim j → + ∞ C ( j, n ) = lim j → + ∞ X j ≥| t | > R ( t, b m ) + R (0 , b m ) ≥ R (0 , b m ) . But R (0 , b m ) is exactly the number of solutions of (57). Hencelim m → + ∞ R (0 , b m ) = + ∞ . Hence when m → ∞ the number of solutions of equation (66) with n = n m goesto infinity. Theorem 10. Assume that ν = ν m is the number of solutions of xyz = b m , z = x + y , x, y, z > , (67)where b m is as defined in Theorem 9. Then under the generalized abc − conjecturewe have k = rad( xyz ) → + ∞ as m → + ∞ andlim m → + ∞ ν m = + ∞ . (68)Further when m → + ∞ , we have the following asymptotic expansion1 ν ν X i =1 z i = k exp " s k log log k (cid:18) k k + O (cid:18) k (cid:19)(cid:19) . (69) References [1]: Robert Olivier, Stewart Cameron L., Tenenbaum Gerald. ”A refinementof the abc conjecture”. Bulletin of the London Mathematical Society.46