# Solutions to Some Classical Diophantine Equations

• Titu Andreescu
• Dorin Andrica
• Ion Cucurezeanu
Chapter

## 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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© 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