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On The Diophantine Equation X2+p2k=yn

Posted on:2012-03-31Degree:MasterType:Thesis
Country:ChinaCandidate:Z W ShangFull Text:PDF
GTID:2120330335963427Subject:Applied Mathematics
Abstract/Summary:
Let f(x) be an irreducible polynomial of degree m≥2 with integral coefficients. Let n≥2 be an integer. Since the work of Siegel, and other mathematicians'contri-bution, we know that the diophantine equation f(x)= yn, in integers x, y has only finitely many solutions, provided that (m, n)≠(2,2). Many papers deal with this particular case ax2+bx+c=dyn, in integers x, y, n≥3.If we already had a=d=1, the equation can be simplified to x2+c=yn, in integers x, y, n≥3.If c< 0, connecting with the unit of a real quadratic field, the diophantine equation became more complicated. So most of recent papers are dealing with such case of c> 0.In this thesis, we assume 101≤p≤1000 is a rational prime and consider the equation x2+p2k=yn(1) with unknown integers x,y,n,k so that x>0,y>1,n≥3 prime, k≥0 and (x,y)= 1. Under the previous assumptions, we give complete solutions of most equations in the form (1), as p runs over all primes between 100 to 1000. Note that the equation (1) is a special case of the diophantine equation x2+C=yn, where C is prime power.
Keywords/Search Tags:Exponential diophantine equations, Lucas numbers, Primitive divisors, Binary quadratic form
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