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 example17 \; \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.
Hi, mist
RăspundețiȘtergere"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.
I am not very expert, but I would like to collaborate, would you let me? I am from Latin America.
RăspundețiȘtergereFor divisibility by 7, the second row is obtained from the first, changing y to -y. I think it's missing
RăspundețiȘtergere7 \mid x+3y\;\Leftrightarrow\;\;7\mid 2x-y\;\Leftrightarrow\;\;7\mid 3x+2y
You are right, I will review this post: https://ogeometrie-cip.blogspot.com/2025/01/anothers-divizibility-relations.html
ȘtergereOn the same model as in your Lemma, I also found six equivalent forms
RăspundețiȘtergere17\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
Okay, so the set of relationships we thought was "complete" isn't like that at all.
ȘtergereIf you want to set records, then I have the following also SIX equivalents
Ștergere17\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
Oh, I see there are SEVEN, so that really is a record !!!
ȘtergereUnfortunately you wrote one thing twice: 17\mid x+4y