Is there a computable upper bound for the height of a solution of a Diophantine equation with a unique solution in positive integers?

Is there a computable upper bound for the height of a solution of a Diophantine equation with a unique solution in positive integers? Apoloniusz Tyszka arxiv:1404.5975v18 [math.lo] 6 Apr 2017 Abstract Let B n ={x i x j = x k, x i + 1= x k : i, j, k {1,...,n}}. For a positive integer n, letξn) denote the smallest positive integer b such that for each systems B n with a unique solution in positive integers x 1,..., x n, this solution belongs to [1, b] n. Let g1)=1, and let gn+1)=2 2gn) for every positive integer n. We conjecture thatξn) g2n) for every positive integer n. We prove:1) the functionξ :N\{0} N\{0} is computable in the limit;2) if a function f :N\{0} N\{0} has a single-fold Diophantine representation, then there exists a positive integer m such that f n)<ξn) for every integer n>m;3) the conjecture implies that there exists an algorithm which takes as input a Diophantine equation Dx 1,..., x p )=0 and returns a positive integer d with the following property: for every positive integers a 1,...,a p, if the tuple a 1,...,a p ) solely solves the equation Dx 1,..., x p )=0in positive integers, then a 1,...,a p d;4) the conjecture implies that if a set M N has a single-fold Diophantine representation, then M is computable;5) for every integer n>9, the inequalityξn)< 2 2n 5 1)2 n 5 + 1 implies that 2 2n 5 + 1 is composite. Key words and phrases: Diophantine equation with a unique solution in positive integers, Fermat prime, single-fold Diophantine representation, upper bound for the height of a solution. 2010 Mathematics Subject Classification: 03D20, 11U05. 1 Introduction In this article, we propose a conjecture which implies that there exists an algorithm which takes as input a Diophantine equation Dx 1,..., x p )=0 and returns a positive integer d with the following property: for every positive integers a 1,..., a p, if the tuple a 1,..., a p ) solely solves the equation Dx 1,..., x p )=0in positive integers, then a 1,...,a p d. Since the height of an integer tuple c 1,..., c p ) is defined as max c 1,..., c p ), our claim asserts that the height of the tuple a 1,..., a p ) does not exceed d. 2 A conjecture on integer arithmetic Let g1)=1, and let gn+1)=2 2gn) for every positive integer n Let B n ={x i x j = x k, x i + 1= x k : i, j, k {1,..., n}}. For a positive integer n, letξn) denote the 1

smallest positive integer b such that for each systems B n with a unique solution in positive integers x 1,..., x n, this solution belongs to [1, b] n. We do not know whether or not there exists a computable functionγ:n\{0} N\{0} such that the inequalityξn) γn) holds for every sufficiently large positive integer n. Theorem 1. The function ξ : N\{0} N\{0} is computable in the limit. Proof. The flowchart in Figure 1 describes an algorithm which computesξn) in the limit. Start Input a positive integer n k := 1 Create a list C of all systems S B n which have exactly one solution in {1,..., k} n Compute the smallest positive integer b {1,..., k} such that each system from C has a solution in {1,..., b} n Print b k := k+1 Fig. 1 An infinite computation ofξn) Lemma 1. For every positive integers b and c, b+1=c if and only if 2 2b 2 2b = 2 2c. Observations 1 and 2 heuristically justify Conjecture 1. Observation 1. For every systems B n, the following new system {x i x j = x k } {22x k = yk : k {1,..., n}} x i x j =x k S x i +1=x k S {y i y i = y k } is equivalent tos. If the systemshas exactly one solution in positive integers x 1,..., x n, then the new system has exactly one solution in positive integers x 1,..., x n, y 1,..., y n. Proof. It follows from Lemma 1. Observation 2. For every positive integer n, the following system x 1 x 1 = x 1 i {1,..., n 1} 2 2x i = x i+1 if n>1) has exactly one solution in positive integers, namely g1),...,gn)). Lemma 1 and Observations 1 2 justify the following conjecture. Conjecture 1. If a systems B n has exactly one solution in positive integers x 1,..., x n, then x 1,..., x n g2n). In other words,ξn) g2n) for every positive integer n. Conjecture 2 generalizes Conjecture 1. Conjecture 2. If a systems B n has only finitely many solutions in positive integers x 1,..., x n, then x 1,..., x n g2n). 2

3 Algebraic lemmas lead to the main theorem LetRng denote the class of all rings K that extendz, and let E n ={1= x k, x i + x j = x k, x i x j = x k : i, j, k {1,..., n}} Th. Skolem proved that every Diophantine equation can be algorithmically transformed into an equivalent system of Diophantine equations of degree at most 2, see [7, pp. 2 3] and [3, pp. 3 4]. The following result strengthens Skolem s theorem. Lemma 2. [10, p. 720]) Let Dx 1,..., x p ) Z[x 1,..., x p ]. Assume that degd, x i ) 1 for each i {1,..., p}. We can compute a positive integer n> p and a systemt E n which satisfies the following two conditions: Condition 1. If K Rng {N,N\{0}}, then x 1,..., x p K D x 1,..., x p )=0 x p+1,..., x n K x 1,..., x p, x p+1,..., x n ) solvest ) Condition 2. If K Rng {N,N\{0}}, then for each x 1,..., x p K with D x 1,..., x p )=0, there exists a unique tuple x p+1,..., x n ) K n p such that the tuple x 1,..., x p, x p+1,..., x n ) solvest. Conditions 1 and 2 imply that for each K Rng {N,N\{0}}, the equation Dx 1,..., x p )=0 and the systemt have the same number of solutions in K. Letα,β, andγdenote variables. Lemma 3. [6, p. 100]) For each positive integers x, y, z, x+y=zif and only if zx+1)zy+1)=z 2 xy+1)+1 Corollary 1. We can express the equation x+y=z as an equivalent systemf, wheref involves x, y, z and 9 new variables, and wheref consists of equations of the formsα+1=γ andα β=γ. Proof. The new 9 variables express the following polynomials: zx, zx+1, zy, zy+1, z 2, xy, xy+1, z 2 xy+1), z 2 xy+1)+1 Lemma 4. Let Dx 1,..., x p ) Z[x 1,..., x p ]. Assume that degd, x i ) 1 for each i {1,..., p}. We can compute a positive integer n> p and a systemt B n which satisfies the following two conditions: Condition 3. For every positive integers x 1,..., x p, D x 1,..., x p )=0 x p+1,..., x n N\{0} x 1,..., x p, x p+1,..., x n ) solvest Condition 4. If positive integers x 1,..., x p satisfy D x 1,..., x p )=0, then there exists a unique tuple x p+1,..., x n ) N\{0}) n p such that the tuple x 1,..., x p, x p+1,..., x n ) solvest. Conditions 3 and 4 imply that the equation Dx 1,..., x p )=0and the systemt have the same number of solutions in positive integers. 3

Proof. Let the systemt be given by Lemma 2. We replace int each equation of the form 1= x k by the equation x k x k = x k. Next, we apply Corollary 1 and replace int each equation of the form x i + x j = x k by an equivalent system of equations of the formsα+1=γ and α β=γ. Lemma 4 implies Theorem 2. Theorem 2. Conjecture 1 implies that there exists an algorithm which takes as input a Diophantine equation Dx 1,..., x p )=0 and returns a positive integer d with the following property: for every positive integers a 1,..., a p, if the tuple a 1,..., a p ) solely solves the equation Dx 1,..., x p )=0 in positive integers, then a 1,..., a p d. Theorem 3. Conjecture 1 implies that there exists an algorithm which takes as input a Diophantine equation Dx 1,..., x p )=0 and returns a positive integer d with the following property: for every non-negative integers a 1,..., a p, if the tuple a 1,..., a p ) solely solves the equation Dx 1,..., x p )=0in non-negative integers, then a 1,..., a p d. Proof. We apply Theorem 2 to the equation Dx 1 1,..., x p 1)=0. 4 Single-fold Diophantine representations The Davis-Putnam-Robinson-Matiyasevich theorem states that every recursively enumerable setm N n has a Diophantine representation, that is a 1,..., a n ) M x 1,..., x m N Wa 1,..., a n, x 1,..., x m )=0 R) for some polynomial W with integer coefficients, see [3]. The polynomial W can be computed, if we know the Turing machine M such that, for all a 1,..., a n ) N n, M halts on a 1,..., a n ) if and only if a 1,..., a n ) M, see [3]. The representationr) is said to be single-fold, if for any a 1,..., a n N the equation Wa 1,..., a n, x 1,..., x m )=0 has at most one solution x 1,..., x m ) N m. The representationr) is said to be finite-fold, if for any a 1,..., a n N the equation Wa 1,..., a n, x 1,..., x m )=0 has only finitely many solutions x 1,..., x m ) N m. Yu. Matiyasevich conjectured that each recursively enumerable setm N n has a single-fold finite-fold) Diophantine representation, see [1, pp. 341 342] and [4, p. 42]. Currently, he seems agnostic on his conjectures, see [5, p. 749]. In [9, p. 581], the author explains why Matiyasevich s conjectures although widely known are less widely accepted. Theorem 4. cf. [8, Theorem 5, p. 711]) Conjecture 1 implies that if a setm N has a single-fold Diophantine representation, then M is computable. In particular, Conjecture 1 contradicts Matiyasevich s conjecture on single-fold Diophantine representations. Proof. Let a setm N have a single-fold Diophantine representation. It means that there exists a polynomial Wx, x 1,..., x m ) with integer coefficients such that b N b M x 1,..., x m N Wb, x 1,..., x m )=0 ) and for every b N the equation Wb, x 1,..., x m )=0 has at most one solution x 1,..., x m ) N m. By Theorem 3, there exists a computable functionθ:n N such that for every b, x 1,..., x m N the equality Wb, x 1,..., x m )=0 implies maxx 1,..., x m ) θb). Hence, we can decide whether or not a non-negative integer b belongs tomby checking whether or not the equation Wb, x 1,..., x m )=0 has an integer solution in the box [0,θb)] m. 4

Observation 3. Theorem 4 remains true if we change the bound g2n) in Conjecture 1 to any other computable boundδn). Theorem 5. If a function f :N\{0} N\{0} has a single-fold Diophantine representation, then there exists a positive integer m such that f n)<ξn) for every integer n>m. Proof. There exists a polynomial Wx 1, x 2, x 3,..., x r ) with integer coefficients such that for each positive integers x 1, x 2, x 1, x 2 ) f x 3,..., x r N\{0} Wx 1, x 2, x 3 1,..., x r 1)=0 and for each positive integers x 1, x 2 at most one tuple x 3,..., x r ) of positive integers satisfies Wx 1, x 2, x 3 1,..., x r 1)=0. By Lemma 4, there exists an integer s 3such that for every positive integers x 1, x 2, x 1, x 2 ) f x 3,..., x s N\{0}Ψx 1, x 2, x 3,..., x s ) E) whereψx 1, x 2, x 3,..., x s ) is a conjunction of formulae of the forms x i + 1= x k and x i x j = x k, the indices i, j, k belong to{1,..., s}, and for each positive integers x 1, x 2 at most one tuple x 3,..., x s ) of positive integers satisfiesψx 1, x 2, x 3,..., x s ). Let [ ] denote the integer part function, and let an integer n is greater than m=2s+2. Then, [ n n + 2] n [ ] n 2 > + s+1 2 and n [ n 2] s 2 0. Let Tn denote the following system with n variables: all equations occurring inψx 1, x 2, x 3,..., x s ) i { 1,..., n [ } n 2] s 2 ui u i = u i t 1 t 1 = t 1 i { 1,..., [ } n 2] 1 ti + 1 = t i+1 t 2 t [ n 2 ] = u u+1 = x 1 if n is odd) t 1 u = x 1 if n is even) x 2 + 1 = y By the equivalencee), the system T n is solvable in positive integers, 2 [ n 2] = u, n= x1, and f n)= f x 1 )= x 2 < x 2 + 1=y Since T n B n and T n has at most one solution in positive integers, y ξn). Hence, f n)<ξn). Corollary 2. If the functionξ :N\{0} N\{0} is dominated by a computable function, then Matiyasevich s conjecture on single-fold Diophantine representations is false. In particular, Conjecture 1 contradicts Matiyasevich s conjecture on single-fold Diophantine representations. 5 Fermat primes Observation 4. Only x 1 = 1 solves the equation x 1 x 1 = x 1 in positive integers. Only x 1 = 1 and x 2 = 2 solve the system{x 1 x 1 = x 1, x 1 + 1= x 2 } in positive integers. For each integer 5

n 3, the following system has a unique solution in positive integers, namely x 1 x 1 = x 1 x 1 + 1 = x 2 i {2,..., n 1} x i x i = x i+1 1, 2, 4, 16, 256,..., 2 2n 3, 2 2n 2). Corollary 3. We have: ξ1)=1 andξ2)=2. The inequalityξn) 2 2n 2 holds for every integer n 3. Primes of the form 2 2n + 1 are called Fermat primes, as Fermat conjectured that every integer of the form 2 2n + 1 is prime [2, p. 1]). Fermat correctly remarked that 2 20 + 1=3, 2 21 + 1=5, 2 22 + 1=17, 2 23 + 1=257, and 2 24 + 1=65537 are all prime [2, p. 1]). It is not known whether or not there are any other Fermat primes [2, p. 206]). Theorem 6. If n N\{0} and 2 2n + 1 is prime, then the following system i {1,..., n} x i x i = x i+1 x 1 + 1 = x n+2 x n+2 + 1 = x n+3 x n+1 + 1 = x n+4 x n+3 x n+5 = x n+4 has a unique solution a 1,..., a n+5 ) in non-negative integers. The numbers a 1,..., a n+5 are positive and max a 1,..., a n+5 )=a n+4 = 2 2n 1 ) 2 n + 1. Proof. The system equivalently expresses that x 1 + 2) x n+5 = x 2n 1 + 1. Since x 2n 1 + 1=x 1 + 2) 2) 2n + 1= 2 n ) 2 2 2n n + 1+x 1 + 2) x 1 + 2) k 1 2) 2n k k k=1 we get 2 n ) 2 n x 1 + 2) x n+5 x 1 + 2) k 1 2) 2n k k = 22n + 1 k=1 Hence, x 1 + 2 divides 2 2n + 1. Since x 1 + 2 2 and 2 2n + 1 is prime, we get x 1 + 2=2 2n + 1 and x 1 = 2 2n 1. Next, x n+1 = x 2n 1 = 2 2n 1 ) 2 n and x n+4 = x n+1 + 1= 2 2n 1)2 n + 1 Explicitly, the whole solution is given by i {1,..., n+1} a i = 2 2n 1 ) 2 i 1 a n+2 = 2 2n a n+3 = 2 2n + 1 a n+4 = 2 2n 1 ) 2 n + 1 ) a n+5 = 1+ 2n 2 n 2 2n + 1 )k 1 2) 2 n k k=1 k 6

Corollary 4. For every integer n>5, if 2 2n 5 + 1 is prime, then In particular, ξ9) ξn) 2 2n 5 1)2 n 5 + 1 )2 9 5 2 29 5 1 + 1= 2 16 1 ) 16 + 1> 2 16 2 15) 16 ) = 2 15 16 = 2 240 > 2 29 2 The numbers 2 2n 5 + 1 are prime when n {6, 7, 8}, but )2 6 5 2 26 5 1 + 1=10<65536=2 26 2 )2 7 5 2 27 5 1 + 1=50626<4294967296=2 27 2 )2 8 5 2 28 5 1 + 1=17878103347812890626< 18446744073709551616= 2 28 2 Corollary 5. For every integer n>9, the inequalityξn)< 2 2n 5 + 1 is composite. 2 2n 5 1)2 n 5 + 1 implies that 6 Two stronger conjectures Conjecture 3 strengthens Conjecture 1. Conjecture 3. If n {3,..., 8}, thenξn)=2 2n 2. If n 9, thenξn) 2 2n 5 1)2 n 5 + 1. Theorem 7. cf. [8, Theorem 2, p. 709]) For each positive integer n, the following system i {1,..., n} x i x i = x i+1 x n+2 + 1 = x 1 x n+3 + 1 = x n+2 x n+3 x n+4 = x n+1 x n+5 x n+5 = x n+5 x n+5 + 1 = x n+6 x n+6 x n+7 = x 1 x n+6 x n+8 = x n+9 x n+9 + 1 = x n+4 7

has a unique solution a 1,..., a n+9 ) in positive integers. The numbers a 1,..., a n+9 satisfy: i {1,..., n+1} a i a n+2 a n+3 = 2+2 2n )2 i 1 = 1+2 2n = 2 2n a n+4 = 1+2 2n 1 ) 2 n a n+5 = 1 a n+6 = 2 a n+7 = 1+2 2n 1 1+2 2 n 1 ) 2 n 1 a n+8 = a n+9 = 1+2 2n 1 ) 2 n 1 Proof. The tuple a 1,..., a n+9 ) consists of positive integers and solves the system. We prove that this solution is unique in positive integers. The equations imply that x 1 is even. The equations imply that x n+4 is odd. The equations x n+5 x n+5 = x n+5 x n+5 + 1 = x n+6 x n+6 x n+7 = x 1 x n+5 x n+5 = x n+5 x n+5 + 1 = x n+6 x n+6 x n+8 = x n+9 x n+9 + 1 = x n+4 i {1,..., n} x i x i = x i+1 x n+2 + 1 = x 1 x n+3 + 1 = x n+2 x n+3 x n+4 = x n+1 imply that x 1 = x n+3 + 2 3 and x 2n 1 = x 1 2) x n+4. This equality and the polynomial identity 2 n 1 x 2n 1 = 2 2n + x 1 2) 1 22n + 2 2n k 1 k x 1 k=1 2 imply that x 1 2 divides 2 2n and x n+4 = x2n 1 x 1 2 = 22n x 1 2 + 1 2n 1 22n + 2 2n 1 k x k 1 } k=1 {{ } the sum of two even numbers as x 1 is even L) Since x 1 3 and x 1 2 divides 2 2n, x 1 = 2+2 k, where k [0, 2 n ] Z. Since x n+4 is odd, the 2 2n equalityl) implies that x 1 2 is odd. Hence, x 1= 2+2 2n. Consequently, x i = a i for every i {1,..., n+9}. 8

Corollary 6. The inequalityξn) 2+2 2n 9) 2 n 9 holds for every integer n 10. Conjecture 4. The equality ξn) = 2+2 2n 9) 2 n 9 integer n. holds for every sufficiently large positive References [1] M. Davis, Yu. Matiyasevich, J. Robinson, Hilbert s tenth problem. Diophantine equations: positive aspects of a negative solution; in: Mathematical developments arising from Hilbert problems ed. F. E. Browder), Proc. Sympos. Pure Math., vol. 28, Part 2, Amer. Math. Soc., Providence, RI, 1976, 323 378, http://dx.doi.org/10.1090/pspum/028.2; reprinted in: The collected works of Julia Robinson ed. S. Feferman), Amer. Math. Soc., Providence, RI, 1996, 269 324. [2] M. Křížek, F. Luca, L. Somer, 17 lectures on Fermat numbers: from number theory to geometry, Springer, New York, 2001. [3] Yu. Matiyasevich, Hilbert s tenth problem, MIT Press, Cambridge, MA, 1993. [4] Yu. Matiyasevich, Hilbert s tenth problem: what was done and what is to be done; in: Hilbert s tenth problem: relations with arithmetic and algebraic geometry Ghent, 1999), Contemp. Math. 270, 1 47, Amer. Math. Soc., Providence, RI, 2000, http://dx.doi.org/10.1090/conm/270. [5] Yu. Matiyasevich, Towards finite-fold Diophantine representations, J. Math. Sci. N. Y.) vol. 171, no. 6, 2010, 745 752, http://dx.doi.org/10.1007%2fs10958-010-0179-4. [6] J. Robinson, Definability and decision problems in arithmetic, J. Symbolic Logic 14 1949), no. 2, 98 114, http://dx.doi.org/10.2307/2266510; reprinted in: The collected works of Julia Robinson ed. S. Feferman), Amer. Math. Soc., Providence, RI, 1996, 7 23. [7] Th. Skolem, Diophantische Gleichungen, Julius Springer, Berlin, 1938. [8] A. Tyszka, A hypothetical way to compute an upper bound for the heights of solutions of a Diophantine equation with a finite number of solutions, Proceedings of the 2015 Federated Conference on Computer Science and Information Systems eds. M. Ganzha, L. Maciaszek, M. Paprzycki); Annals of Computer Science and Information Systems, vol. 5, 709 716, IEEE Computer Society Press, 2015, http://dx.doi.org/10.15439/2015f41. [9] A. Tyszka, All functions g: N N which have a single-fold Diophantine representation are dominated by a limit-computable function f :N\{0} N which is implemented in MuPAD and whose computability is an open problem; in: Computation, cryptography, and network security eds. N. J. Daras and M. Th. Rassias), Springer, Cham, Switzerland, 2015, 577 590,http://dx.doi.org/10.1007/978-3-319-18275-9_24. 9

[10] A. Tyszka, Conjecturally computable functions which unconditionally do not have any finite-fold Diophantine representation, Inform. Process. Lett. 113 2013), no. 19 21, 719 722, http://dx.doi.org/10.1016/j.ipl.2013.07.004. Apoloniusz Tyszka University of Agriculture Faculty of Production and Power Engineering Balicka 116B, 30-149 Kraków, Poland Email: rttyszka@cyf-kr.edu.pl 10