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Small Prime Solutions to Cubic Diophantine Equations

Published online by Cambridge University Press:  20 November 2018

Zhixin Liu*
Affiliation:
Department of Mathematics, School of Science, Tianjin University, Tianjin 300072, P. R. China e-mail: [email protected]
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Abstract.

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Let ${{a}_{1}},...,{{a}_{9}}$ be nonzero integers and $n$ any integer. Suppose that ${{a}_{1}}+\cdot \cdot \cdot +{{a}_{9}}\,\equiv \,n\,\left( \bmod \,2 \right)$ and $\left( {{a}_{i}},\,{{a}_{j}} \right)\,=\,1$ for $1\,\le \,i\,<\,j\,\le \,9$. In this paper we prove the following:

(i) If ${{a}_{j}}$ are not all of the same sign, then the cubic equation ${{a}_{1}}p_{1}^{3}\,+\cdot \cdot \cdot +\,{{a}_{9}}\,p_{9}^{3}\,=\,n$ has prime solutions satisfying ${{p}_{j}}\,\ll \,{{\left| n \right|}^{{1}/{3}\;}}\,+\,\max {{\left\{ \left| {{a}_{j}} \right| \right\}}^{14+\varepsilon }}$.

(ii) If all ${{a}_{j}}$ are positive and $n\,\gg \,\max {{\left\{ \left| {{a}_{j}} \right| \right\}}^{43+\varepsilon }}$, then ${{a}_{1}}p_{1}^{3}\,+\cdot \cdot \cdot +\,{{a}_{9}}\,p_{9}^{3}\,=\,n$ is solvable in primes ${{p}_{j}}$.

These results are an extension of the linear and quadratic relative problems.

Type
Research Article
Copyright
Copyright © Canadian Mathematical Society 2013

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