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
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© 2009 Birkhäuser Boston
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Andreescu, T., Andrica, D., Cucurezeanu, I. (2009). Solutions to Some Classical Diophantine Equations. In: An Introduction to Diophantine Equations. Birkhäuser Boston. https://doi.org/10.1007/978-0-8176-4549-6_6
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DOI: https://doi.org/10.1007/978-0-8176-4549-6_6
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