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Let there be a linear congruence $a+b y \equiv 0 \pmod{m}$, with $y$ and $m$ ($m$ is a prime) values known. Do all the integer $(a,b)$ pairs satisfying the congruence form a lattice? If yes, how can I find two vectors (i.e. the basis) which generate the lattice mentioned?

vonbrand
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user2204800
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1 Answers1

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Hint $\rm\,\ mod\ m\!:\ a\equiv -by \iff (a,b) = (km-ny,n)\, =\, k(m,0) + n(-y,1)$

Math Gems
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  • thanks a lot. So, the same lattice is generated by k(m,0)+n(m−y,1) and also by k(m,0)+n((−y)%m,1). Am I right? – user2204800 Mar 26 '13 at 13:04
  • Yes, $\rm\ mod\ (m,0)!:\ (-y,1),\equiv, (-y,1)+j(m,0), =, (jm!-!y,1),\ $ therefore $\rm:(m,0),\Bbb Z + (-y,1),\Bbb Z, =, (m,0),\Bbb Z + (jm!-!y,1),\Bbb Z\ \ $ – Math Gems Mar 26 '13 at 13:45