Does Every Recursively Enumerable Set Admit a Finite-Fold Diophantine Representation?

When a number is to be multiplied by itself repeatedly,  the result is a power    of this number.

• In the third paper (MJCNT, 2013), for each (irreducible) binary form attached to an algebraic number field, which is not a totally real cubic field, we exhibited an infinite family

The general case of Schmidt’s Subspace Theorem ([1], Theorem 2.5) involves a finite set of places of a number field K, containing the places at infinity, and instead of |x| −

The following result shows that if the interval width determined by the function ~ is sufficiently large then the Kaufman measure of Eq(r is essentially equal to the

There is a motivation from logic : Were Z ⊂ Q diophantine, then Hilbert’s 10th problem over the rationals would have a negative answer (using Matijasevich’s theorem over Z).. Let

If one could prove an analogue of Theorem 1 for the Gauss map then this would imply by Proposition 3 the result dim C(r) == 2/r. At present we have not been able to prove such

In the conclusions, we briefly discuss the autonomous significance of our specifi- cation independently of Jones’s conjecture, and address the issue of determining a lower bound for

One way out of this difficulty was indicated in [2], and has been recently recalled in [6, 9]: If one managed to prove that there are only a finite number of solutions to a