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joi, 9 aprilie 2020

SIMULTANEOUS DIVISIBILITY of 1-FORMS

          Is about Problem 1 from Eőtvős Mathematical Competition 1894.
In Romanian there is this book: "Probleme de matematica pentru concurs";
the book can be downloaded from here (the solution is on the pages 23-28).

          The stattement of the problem can be written as a logical equivalence

(1)                                      17 \; \mid 2x+3y \; \; \Leftrightarrow \; 17 \; \mid 9x+5y.

          We will give a list of equivalences, "completely" in some sense, from which we can deduce (almost) any of such a relations. 

LEMA For any integers x, y the following equivalences hold:

17 \; \mid \; I(x,y)=2x+3y \; \Leftrightarrow \; 17 \; \mid \;II(x,y)=x-7x \; \Leftrightarrow \; 17 \; \mid \; III(x,y)=3x-4y \; \Leftrightarrow

\Leftrightarrow \;17 \; \mid \; IV(x,y)=5x-y \; \Leftarrow \; 17 \; \mid V(x,y)=7x+2y \; \Leftarrow \; 17 \; \mid \; VI(x,y)=8x-5y.


Proof. We have 17 \; \mid 2x+3y \; \Leftrightarrow \; \widehat{2}x+\widehat{3}y=\widehat{0}, where \widehat{m} mean the congruence class modulo 17 determined by m; their set is \mathbb{Z}/17\mathbb{Z} (actually it is the field GF(17)). If we multiply last equation by \widehat{9} and we take into account that, in GF(17),
 \widehat{9} \cdot \widehat{2}=\widehat{1}, \; \widehat{9} \cdot \widehat{3}=\widehat{10}=-\widehat{7}
we obtain \widehat{1}x-\widehat{7}y=\widehat{0} \; \Leftrightarrow \; 17 \; \mid \; x-7y.
          In the same way, starting with \widehat{2}x+\widehat{3}y=\widehat{0} and multiplyng it by \widehat{10}, \widehat{11}, \widehat{12}, \widehat{4} respectively, we get the remained equivalences.
\blacksquare

          Apparently we didn't find the desired result. It comes if we were to multiply \widehat{2}x+\widehat{3y}=\widehat{0} by \widehat{13} and taking into account that
 \widehat{13} \cdot \widehat{2}=\widehat{9} and \widehat{13} \cdot \widehat{3}=\widehat{5} 
but patience \cdot \cdot \cdot The word "completely" means that no other divisibility relation that can be obtained from 17 \mid 2x+3y it must be able to be expressed by ones of the forms I-VI (and apliying other simple divisibility properties). From example
17 \; \mid I(x,y) \; \stackrel{LEMA} \Leftrightarrow  17 \; \mid VI(x,y) \Leftrightarrow 17 \; \mid 17x-VI(x,y)=9x+5y.

          Finally, we conclude anothers equivalences that we will use in a similar problem:


7\; \mid x+2y \; \Leftrightarrow \; 7 \; \mid 2x-3y \; \Leftrightarrow \; 7 \; \mid 3x-y;


7\; \mid 3x+y \; \Leftrightarrow \; 7\; \mid x-2y \; \Leftrightarrow \; 7 \; \mid 2x+3y.

8 comentarii:

  1. Hi, mist
    "Finally"" in your last three rows contain redundant affirmations (I.e. more than necessary)
    Is enoug to change "y" in first row of equivalences with "-y" in order to obtain the second row of equivalences (or vice versa).
    Oh, dear, if we multiply x+2y with ^2 , ^3 respectively, we obtain all of 1-forms needed for a "completely" list of congruences.

    RăspundețiȘtergere
  2. I am not very expert, but I would like to collaborate, would you let me? I am from Latin America.

    RăspundețiȘtergere
  3. For divisibility by 7, the second row is obtained from the first, changing y to -y. I think it's missing

    7 \mid x+3y\;\Leftrightarrow\;\;7\mid 2x-y\;\Leftrightarrow\;\;7\mid 3x+2y

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    Răspunsuri
    1. You are right, I will review this post: https://ogeometrie-cip.blogspot.com/2025/01/anothers-divizibility-relations.html

      Ștergere
  4. On the same model as in your Lemma, I also found six equivalent forms
    17\mid 2x+3y \Leftrightarrow 17\mid 3x-4y \Leftrightarrow 17 \mid 5x-y \Leftrightarrow
    \Leftrightarrow 17 \mid 7x+2y \Leftrightarrow 17\mid x-7y \Leftrightarrow 17\mid 8x-5y

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    Răspunsuri
    1. Okay, so the set of relationships we thought was "complete" isn't like that at all.

      Ștergere
    2. If you want to set records, then I have the following also SIX equivalents
      17\mid 4x-y\Leftrightarrow 17\mid 5x+3y \Leftrightarrow 17\mid x+4y \Leftrightarrow 17\mid 3x-5y \Leftrightarrow
      \Leftrightarrow 17\mid 7x-6y \Leftrightarrow 17\mid 6x+7y \Leftrightarrow 17\mid x+4y

      Ștergere
    3. Oh, I see there are SEVEN, so that really is a record !!!
      Unfortunately you wrote one thing twice: 17\mid x+4y

      Ștergere