I am asking if it is possible to choose the following case: $λ_{2}<λ_{5}$ and $λ_{2}λ_{6}>0$

Question

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

$$λ_{1}A+λ_{2}B=1$$

$$λ_{3}A+λ_{4}C=1$$

$$λ_{5}B+λ_{6}C=1$$

I know that by the Bézout theorem (https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity#Structure_of_solutions) that there exists an infinity of them.

I am asking if it is possible to choose the following case: $λ_{2}<λ_{5}$ and $λ_{2}λ_{6}>0$ and $$λ_{3}>(-λ_{1}λ_{5})/(λ_{2}-λ_{5})$$ and $$λ_{3}>(-λ_{1}λ_{4}λ_{5})/(λ_{2}λ_{6})$$

If, $A,B,C > 0$, yes: add $pB$ to $\lambda_1$, substract $pB$ from $\lambda_2$, add $pC$ to $\lambda_5$ and substract $pB$ from $\lambda_6$, with $p$ large enough. — Aphelli, Dec 15 '20 at 18:20

Aphelli · Answer 1 · 2020-12-15T22:17:19.667

1

I'm paraphrasing my comment.

If, $A,B,C > 0$, yes: add $pB$ to $\lambda_1$, substract $pB$ from $\lambda_2$, add $pC$ to $\lambda_5$ and substract $pB$ from $\lambda_6$, with $p$ large enough.

For the edited version (with the first added condition), yes, you add $pC$ to $\lambda_3$ and take $pA$ from $\lambda_4$.

edited Dec 15 '20 at 22:17

answered Dec 15 '20 at 18:25

Aphelli

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I am asking if it is possible to choose the following case: $λ_{2}<λ_{5}$ and $λ_{2}λ_{6}>0$

1 Answers1