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

$$λ_{1}A+λ_{2}B=1$$

$$λ_{3}A+λ_{4}C=1$$

$$λ_{5}B+λ_{6}C=1$$

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

$$A=-(λ₂λ₄-λ₂λ₆-λ₄λ₅)/(λ₁λ₄λ₅+λ₂λ₃λ₆)$$

$$B= (λ₁λ₄-λ₁λ₆+λ₃λ₆)/(λ₁λ₄λ₅+λ₂λ₃λ₆)$$

$$C= (λ₂λ₃+λ₁λ₅-λ₃λ₅)/(λ₁λ₄λ₅+λ₂λ₃λ₆)$$

Then I am asking if there is any problem with representing $A,B,C$ in terms of $λ_{1},λ_{2},λ_{3},λ_{4},λ_{5},λ_{6}$. In particular, how one can guaranty that $λ₁λ₄λ₅+λ₂λ₃λ₆$ is not a zero.

Safwane
  • 3,840

1 Answers1

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There's no guarantee that it will not be zero.

Example: 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 $16.5.9=9.8.10$.

Chrystomath
  • 10,798
  • But if there exist some lamds verifying $(λ₁λ₄λ₅+λ₂λ₃λ₆)$ is not zero and $(λ₂λ₄-λ₂λ₆-λ₄λ₅)(λ₁λ₄λ₅+λ₂λ₃λ₆)>0$, then the expression $-(λ₂λ₄-λ₂λ₆-λ₄λ₅)/(λ₁λ₄λ₅+λ₂λ₃λ₆)$ is must be a representation of $A$. – Safwane Dec 22 '20 at 11:26