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Solutions to Some Classical Diophantine Equations

  • Titu AndreescuEmail author
  • Dorin Andrica
  • Ion Cucurezeanu
Chapter
  • 1.7k Downloads

Abstract

Solution. Working modulo 3, we have y ≡ 1 (mod 3); hence y = 1 + 3s, s ∈ ℤ. The equation becomes 6x − 15z = −9 − 30s, or equivalently, 2x − 5z = −3 − 10s. Passing to modulo 2 yields z ≡ 1 (mod 2), i.e., z = 1 + 2t, t ∈ ℤ and x = 1 -5s + 5t. Hence the solutions are
$$(x,y,z)=(1 -5s + 5t,1 + 3s,1 +2t),\quad s,t \in \mathbb{Z}.$$

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Copyright information

© Birkhäuser Boston 2009

Authors and Affiliations

  • Titu Andreescu
    • 1
    Email author
  • Dorin Andrica
    • 2
    • 3
  • Ion Cucurezeanu
    • 4
  1. 1.School of Natural Sciences and MathematicsUniversity of Texas at DallasRichardsonUSA
  2. 2.Faculty of Mathematics and Computer ScienceBabeş-Bolyai UniversityCluj-NapocaRomania
  3. 3.Department of Mathematics College of ScienceKing Saud UniversityRiyadhSaudi Arabia
  4. 4.Faculty of Mathematics and Computer ScienceOvidius University of ConstantaConstantaRomania

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