| J. Maurice Rojas, Uncomputably large integral points on algebraic plane curves?, Theoretical Computer Science, this issue. |
....of infinitely many integral points, there is an underlying (additive or multiplicative) group action which can be used to describe the distribution of the integral points. In this note we will prove two theorems illustrating this last statement. As a corollary, we will prove a conjecture of Rojas [3] concerning the distribution of integral points relative to the real topology. Little of the material in this note will be new to the experts, but we hope that an elementary exposition will be a useful addition to the literature. Remark. A classical work of Runge [4] gives a necessary condition ....
....C(Z) is an infinite set, then the closure of C(Z) in C(R) for the real topology) is equal to C1 . In other words, if #C(Z) 1, then every real point at infinity is an accumulation point of C(Z) As a special case of Theorem B, we obtain an affirmative answer to Conjecture 1 in Section 6 of [3]. Corollary C. Suppose C ae C 2 is a curve defined over Z and irreducible over C . Suppose further that C(Z) is an infinite set and that some irreducible component of C(R) has non compact intersection with the first quadrant. Then C contains infinitely many positive integral points, that is, ....
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J. Maurice Rojas, Uncomputably large integral points on algebraic plane curves?, Theoretical Computer Science, this issue.
....of detecting a root in N 2 , Z 2 , or even Q 2 , for an arbitrary polynomial in Z[x 1 ; x 2 ] is still completely open. 1 In particular, the major solved special cases so far have only extremely ineffective complexity and height bounds. See, e.g. the introduction and references of [Roj99a]. 1.1 Dimension Zero We will thus go one dimension lower 2 in order to prove a useful result. Main Theorem 1 Suppose F : f 1 ; f m ) is a system of polynomials in Z[x 1 ; x n ] and let Z be the zero set of F in C n . Assume further that dimZ0. Then: 1. We can find all ....
....and Julia Robinson have shown [MR74, Jon81] that sentences of the form 9u 9v 8x 9y f(u; v; x; y) 0, for arbitrary input f 2Z[u; v; x; y] are already undecidable. However, the decidability of sentences of the form 9v 8x 9y f(v; x; y) 0 was an open question until recently: In [Roj99a] it was shown that these sentences can be decided by a Turing machine, once the input f is suitably restricted. Roughly speaking, deciding the prefix 989 is equivalent to determining whether an algebraic surface has a slice (parallel to the (x; y) plane) densely peppered with positive integral ....
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Rojas, J. Maurice, "Uncomputably Large Integral Points on Algebraic Plane Curves?," Theoretical Computer Science, special issue in honor of Manuel
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