The Beckman-Quarles theorem for continuous mappings from C^n to C^n

Let varphi_n:C^n times C^n->C, varphi_n((x_1,...,x_n),(y_1,...,y_n))=sum_{i=1}^n (x_i-y_i)^2. We say that f:C^n->C^n preserves distance d>=0, if for each X,Y in C^n varphi_n(X,Y)=d^2 implies varphi_n(f(X),f(Y))=d^2. We prove: if n>=2 and a continuous f:C^n->C^n preserves unit distance, then f has a form I circ (rho,...,rho), where I:C^n->C^n is an affine mapping with orthogonal linear part and rho:C->C is the identity or the complex conjugation. For n >=3 and bijective f the theorem follows from Theorem 2 in [8].Comment: 10 pages, LaTeX2e, the version which appeared in Aequationes Mathematica

A hypothetical upper bound on the heights of the solutions of a Diophantine equation with a finite number of solutions

Let f(1)=1, and let f(n+1)=2^{2^{f(n)}} for every positive integer n. We conjecture that if a system S \subseteq {x_i \cdot x_j=x_k: i,j,k \in {1,...,n}} \cup {x_i+1=x_k: i,k \in {1,...,n}} has only finitely many solutions in non-negative integers x_1,...,x_n, then each such solution (x_1,...,x_n) satisfies x_1,...,x_n \leq f(2n). We prove: (1) the conjecture implies that there exists 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, (2) the conjecture implies that the question whether or not a Diophantine equation has only finitely many rational solutions is decidable with an oracle for deciding whether or not a Diophantine equation has a rational solution, (3) the conjecture implies that the question whether or not a Diophantine equation has only finitely many integer solutions is decidable with an oracle for deciding whether or not a Diophantine equation has an integer solution, (4) the conjecture implies that if a set M \subseteq N has a finite-fold Diophantine representation, then M is computable.Comment: 13 pages, section 7 expande