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Conjecturally computable functions which unconditionally do not have any finite-fold Diophantine representation

机译:无条件不具有任何有限倍丢番图表示形式的猜想可计算函数

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摘要

We define functions f_1, f_2, g_1, g_2: N {0} → N {0} and prove that they do not have any finite-fold Diophantine representation. We conjecture that if a system S (≤) {x_i + x_j = x_k. x_i · x_j = x_k: i, j , k (≤) {1,...,n}} has only finitely many solutions in positive integers x_1,...,x_n, then each such solution (x_1,...,x_n) satisfies x_1,..., x_n (≤) 2~(2~(n-1)). Assuming the conjecture, we prove: (1) the functions f_1, f_2, g_1 g_2 are computable, (2) there is an algorithm which takes as input a Diophantine equation, returns an integer, and this integer is greater than the heights of integer (non-negative integer, positive integer, rational) solutions if the solution set is finite, (3) a finite-fold Diophantine representation of the function N (≥) n→2~n (≥) N does not exist, (4) if a set M (∈) N is recursively enumerable but not recursive, then a finite-fold Diophantine representation of M does not exist.
机译:我们定义函数f_1,f_2,g_1,g_2:N {0}→N {0}并证明它们不具有任何有限倍的Diophantine表示。我们推测,如果系统S(≤){x_i + x_j = x_k。 x_i·x_j = x_k:i,j,k(≤){1,...,n}}仅具有有限个正整数x_1,...,x_n的解,然后每个这样的解(x_1,... ,x_n)满足x_1,...,x_n(≤)2〜(2〜(n-1))。假设该猜想,我们证明:(1)函数f_1,f_2,g_1 g_2是可计算的,(2)有一种算法以Diophantine方程为输入,返回一个整数,并且该整数大于整数的高度(非负整数,正整数,有理数)解,如果解集是有限的,(3)函数N(≥)n→2〜n(≥)N的有限倍Diophantine表示不存在,(4 )如果集合M(∈)N可递归枚举但不可递归,则M的有限倍丢番图表示不存在。

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