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All numbers here are integers. If $gcd(a,b)=1$ then if $(x_0,y_0)\neq(x_1,y_1)$ with $0<x_i<b$ and $0<y_i<a$ we have $ax_0+by_0\neq ax_1+by_1$.

Similarly is it true $(x_0,y_0,z_0)\neq(x_1,y_1,z_1)$ with $0<x_i<b$, $0<y_i<\min(a,b)$ and $0<z_i<a$ we will have $a^2x_0+aby_0+b^2z_0\neq a^2x_1+aby_1+b^2z_1$?

  • This is hard to read. Why can't I take $x_0=b$, $y_0=0$, $x_1=0$, $y_1=a$? Then it looks like we get $ab$ on both sides. Or does $x_i\in (0,b)$ refer to the open interval $0<x_i<b$? If so take $x_0=\frac b2$, $y_1=\frac a2$. – lulu May 29 '16 at 12:18
  • $0\notin(0,a)$.All numbers here are integers. –  May 29 '16 at 12:19
  • So your condition reads $0<x_i<b$ etc.? But then it seems trivial: take a supposed counterexample and rewrite your equality as $a(x_0-x_1)=b(y_1-y_0)$. Since $\gcd (a,b)=1$ we see that $a$ divides $y_1-y_0$ but that is plainly impossible. – lulu May 29 '16 at 12:21
  • what you state I have already written in first paragraph. –  May 29 '16 at 12:25
  • Still seems trivial. rewrite to get $a^2(x_0-x_1)+ab(y_0-y_1)+b^2(z_0-z_1)=0$ so $b$ divides $x_0-x_1$ and $a$ divides $z_0-z_1$ – lulu May 29 '16 at 12:37
  • oops ok......!! –  May 29 '16 at 18:26
  • @lulu should we have $0<x_i<b$, $0<y_i<\min(a,b)$ and $0<z_i<a$. Even $0\leq x_i\leq b$, $0\leq y_i\leq \min(a,b)$ and $0\leq z_i\leq a$ works right? –  May 30 '16 at 09:19
  • No...take $a=3,b=4$ and ${x_0,y_0,z_0}={4,1,0}$ , ${x_1,y_1,z_1}={0,0,3}$. (there are infinitely many counterexamples of this form). – lulu May 30 '16 at 10:59
  • but if one of $0$ or $b$ is avoided possible? you are using both $4$ and $0$? may be avoiding $0$ will help? –  May 30 '16 at 11:00
  • Well, there are counterexamples of the form ${x_0,y_0,z_0}={X,b,0}$, ${x_1,y_1,z_1}={X,0,a}$ so you need to avoid those as well. If you play around with the constraints I expect you can find some that work...surely, though, context should tell you what constraints you actually need? – lulu May 30 '16 at 11:42
  • @lulu yes but you are still using $0$. If you avoid $0$ may be you wont get match. –  May 30 '16 at 11:45
  • Well, the fact that, say, $b|(x_0-x_1)$ is very strong. If you can block the solution $x_0-x_1=b$ then you quickly rule out everything non-trivial. – lulu May 30 '16 at 11:46
  • @luly ahha then $x_0=x_1=b$ not allowed then you wont get match even with $0$ allowed. –  May 30 '16 at 11:49
  • Not following. You need at least one of $(x_0-x_1)$ and $(z_0-z_1)$ to be non-zero. Divisibility then pretty much tells you what values they have to be...if you block those out by some constraint or other then there are no non-trivial solutions. – lulu May 30 '16 at 11:52
  • @lulu do you know this http://math.stackexchange.com/questions/1805533/counting-solutions-to-equation-summing-to-0? –  May 30 '16 at 11:59
  • I'll take a look later. – lulu May 30 '16 at 12:00

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