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Let $A,B,C$ be three pairiwise coprime positive integers, i.e., there exist six integers $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5},\lambda_{6}$ such that

$$\lambda_{1}A+\lambda_{2}B=1$$

$$\lambda_{3}A+\lambda_{4}C=1$$

$$\lambda_{5}B+\lambda_{6}C=1$$

Solving with respect to $A,B,C$, we get

$$A=-(\lambda_{2}\lambda_{4}-\lambda_{2}\lambda_{6}-\lambda_{4}\lambda_{5})/(\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6})$$

$$B= (\lambda_{1}\lambda_{4}-\lambda_{1}\lambda_{6}+\lambda_{3}\lambda_{6})/(\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6})$$

$$C= (\lambda_{2}\lambda_{3}+\lambda_{1}\lambda_{5}-\lambda_{3}\lambda_{5})/(\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6})$$

Generally, it is not possible to decide if $\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6}$ is not zero and one can find examples about this.

My question:

If there exist some $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5},\lambda_{6}$ verifying $(\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6})$ is not zero and $(\lambda_{2}\lambda_{4}-\lambda_{2}\lambda_{6}-\lambda_{4}\lambda_{5})(\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6})<0$, then show that the expression $-(\lambda_{2}\lambda_{4}-\lambda_{2}\lambda_{6}-\lambda_{4}\lambda_{5})/(\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6})$ must be a representation of the positive integer $A$.

Jair Taylor
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Safwane
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  • $\lambda_i=1$ contradicts your question. Did I miss something? – Zerox Dec 22 '20 at 12:01
  • @Zerox: Yes. the lamds are not arbitrary, they are a result of Bézout theorem and you have infinitely of them. – Safwane Dec 22 '20 at 12:02
  • from ypour earlier question, answer by @Chrystomath Let $A=2\times11$, $B=3\times13$, $C=5\times7$. Then $$\begin{matrix}16A-9B=1\8A-5C=1\9B-10C=1\end{matrix}$$ yet $16.5.9=9.8.10$. – Will Jagy Dec 22 '20 at 17:06
  • @WillJagy: All the question depends on the main assumption that the determinant is not a zero among other conditions. – Safwane Dec 22 '20 at 17:09
  • good. Again, have you taken a class in linear algebra? Your question is a standard matrix construction – Will Jagy Dec 22 '20 at 17:10
  • @WillJagy: The question is about the fact that the positive integer can be represented by rational numbers. – Safwane Dec 22 '20 at 17:18

1 Answers1

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$$ M = \left( \begin{array}{ccc} 0 & \lambda_2 & \lambda_1 \\ \lambda_4 & 0 & \lambda_3 \\ \lambda_6 & \lambda_5 & 0 \\ \end{array} \right) $$ in $$ \left( \begin{array}{ccc} 0 & \lambda_2 & \lambda_1 \\ \lambda_4 & 0 & \lambda_3 \\ \lambda_6 & \lambda_5 & 0 \\ \end{array} \right) \left( \begin{array}{c} C\\ B \\ A \\ \end{array} \right) = \left( \begin{array}{c} 1\\ 1 \\ 1 \\ \end{array} \right) $$

$$ M^{-1} = \frac{1}{ \lambda_6 \lambda_3 \lambda_2 + \lambda_5 \lambda_4 \lambda_1} \hspace{2mm} \left( \begin{array}{ccc} - \lambda_5 \lambda_3 & \lambda_5 \lambda_1 & \lambda_3 \lambda_2 \\ \lambda_6 \lambda_3 & - \lambda_6 \lambda_1 & \lambda_4 \lambda_1 \\ \lambda_5 \lambda_4 & \lambda_6 \lambda_2 & - \lambda_4 \lambda_2 \\ \end{array} \right) $$

Indeed, it is routine to confirm the matrix product $$ \left( \begin{array}{ccc} 0 & \lambda_2 & \lambda_1 \\ \lambda_4 & 0 & \lambda_3 \\ \lambda_6 & \lambda_5 & 0 \\ \end{array} \right) \left( \begin{array}{ccc} - \lambda_5 \lambda_3 & \lambda_5 \lambda_1 & \lambda_3 \lambda_2 \\ \lambda_6 \lambda_3 & - \lambda_6 \lambda_1 & \lambda_4 \lambda_1 \\ \lambda_5 \lambda_4 & \lambda_6 \lambda_2 & - \lambda_4 \lambda_2 \\ \end{array} \right) = \left( \begin{array}{ccc} \lambda_6 \lambda_3 \lambda_2 + \lambda_5 \lambda_4 \lambda_1 & 0 & 0 \\ 0 & \lambda_6 \lambda_3 \lambda_2 + \lambda_5 \lambda_4 \lambda_1 & 0 \\ 0 & 0 & \lambda_6 \lambda_3 \lambda_2 + \lambda_5 \lambda_4 \lambda_1 \\ \end{array} \right) $$

This tells us that $$ \left( \begin{array}{c} C\\ B \\ A \\ \end{array} \right) = \frac{1}{ \lambda_6 \lambda_3 \lambda_2 + \lambda_5 \lambda_4 \lambda_1} \hspace{2mm} \left( \begin{array}{ccc} - \lambda_5 \lambda_3 & \lambda_5 \lambda_1 & \lambda_3 \lambda_2 \\ \lambda_6 \lambda_3 & - \lambda_6 \lambda_1 & \lambda_4 \lambda_1 \\ \lambda_5 \lambda_4 & \lambda_6 \lambda_2 & - \lambda_4 \lambda_2 \\ \end{array} \right) \left( \begin{array}{c} 1\\ 1 \\ 1 \\ \end{array} \right) $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

hello world

Will Jagy
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