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Let $ax+by=n<(a-1)(b-1)$. This has exactly $(a-1)(b-1)/2$ non-negative solutions.

I understand that $ax+by\geq (a-1)(b-1)+1$ has a non-negative solution because the spacing between solution is greater than $\sqrt{a^2+b^2}$. However, in the situation $ax+by<(a-1)(b-1)$, I'm not sure how to show this has $(a-1)(b-1)/2$ non-negative solutions. I thought this would have no solutions since it would seem like the points are too close.

emka
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  • http://math.stackexchange.com/questions/62048/determining-number-of-positive-integer-solutions-to-ax-by-cz-dw-z and http://www.iaeng.org/IJAM/issues_v40/issue_1/IJAM_40_1_01.pdf – lab bhattacharjee Feb 01 '14 at 09:28
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    @labbhattacharjee I don't think that's what he means. Probably he's looking for the number of integers in $[0,,(a-1)(b-1)-1]$ that can be written as a linear combination with positive coefficients. In that case, Morris, here's my hint: try to show that $c$ has a solution if and only if $(a-1)(b-1)-1-c$ has not. – Bart Michels Feb 01 '14 at 10:31
  • I'm not sure I follow. – emka Feb 01 '14 at 10:43

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