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I'm trying to solve the following question.

Let $(b_1, b_{2}, b_3) \in \mathbb{Z}^3$ such that $0 < b_1 \leq b_{2} < b_3$ and $\gcd\{b_3, b_i\}= 1$ for $i=1, 2$.

Let $Q(b_1, b_{2}, b_3)$ be the convex hull of ${a_1 e_1 + a_{2} e_{2} + a_3 (b_1, b_{2}, b_3)}$ where ${e_1, e_2, e_3}$ is the standard basis of $\mathbb{R}^3$ and $a_i \in \{0, 1\}$ for $i=1, 2, 3$.

Does there exist a vector $(c_1, c_2, c_3) \in Q(b_1, b_2, b_3) \cap \mathbb{Z}^3$ such that $\gcd\{c_1, c_3\} = 1$, $\gcd\{c_2, c_3\} = 1$ and $\gcd\{c_2b_3 - b_2 c_3, -(c_1b_3 - b_1 c_3), c_2b_1 - c_1b_2\} = 1$?

My approach is the following.

Let $z=i$ be the plane in $\mathbb{R}^3$. Since $\{z =i\} \cap Q(b_1, b_2, b_3)$ is a parallel translate of the standard square in $\mathbb{R}^2 \times 0$ for $i= 1, \ldots, b_3-1$, the intersection $\{z =i\} \cap Q(b_1, b_2, b_3)$ contains an element $(u_1^i, u_2^i, u_3^i) \in \mathbb{Z}^3$ for $i=1, \ldots, b_3-1$. If $i=1$, then $(u_1^1, u_2^1, u_3^1)$ is either $(0,1,1)$ or $(1, 1,1)$. If $(0,1,1)$ and $(1, 1,1)$ do not satisfy the question, then I try to show for $i=2$. If the question and approach are not clear please let me know.

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Where did this question arise? – David Roberts Feb 10 '14 at 9:18
If I read things correctly, (0,1,1) is never in the convex hull. – The Masked Avenger Feb 10 '14 at 16:28
Also, if you allow b1=b2, I can imagine a situation where all the c's in the lattice points inside the convex hull (except for four corners) satisfy c1=c2, which makes chances slim for your complicated gcd to be satisfied. – The Masked Avenger Feb 10 '14 at 16:39
In particular, I think the answer is no for your b being (5,5,42). – The Masked Avenger Feb 10 '14 at 16:44
@ The Masked: Thank you very much for your answer. To make the answer yes when b1=b2, I considered the case (c1, c2, c3) =(0,1,1) which is never in the convex hull (as you observed). So I need to prove when b1 < b2 < b3. – Sintu Feb 10 '14 at 17:23

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