Other divisibility relations of 1-forms

               In post from April 9, 2020 I stated divisibility relations by 7 of some 1-forms. They are incomplete (even redundant), and should look like this:

$$7\mid\;x+2y\;\;\Leftrightarrow\;\;7\mid\;2x-3y\;\;\Leftrightarrow\;\;7\mid\;3x-y\; ;\tag{1}$$

$$7\mid\;x+3y\;\;\Leftrightarrow\;\;7\mid\;2x-y\;\;\Leftrightarrow\;\;7\mid\;3x+2y\;. \tag{2}$$

If in (1) we write $-y\;$ instead of $\;y\;$ we get

$$7\mid\;x-2y\;\;\Leftrightarrow\;\;7\mid\;2x+3y\;\;\Leftrightarrow\;\;7\mid\;3x+y \tag{1a}$$

that is, the second divisibility from the old post. So (1) and (1a) express the same thing. If in (2) we write $-y\;$ instead of $\;y\;$ we get

$$7\mid\;x-3y\;\;\Leftrightarrow\;\;7\mid\;2x+y\;\;\Leftrightarrow\;\;7\mid\;3x-2y. \tag{2a}$$

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          Suggested by a problem elsewhere, we will state a set of divisibility equivalent to $13\mid 3x-2y.$ I got it for this:

$$13\mid 3x-2y\;\Leftrightarrow\;13\mid x-5y\;\Leftrightarrow\;13\mid 2x+3y\;\Leftrightarrow\;13 \mid 5x+y \tag{13a}$$

and by replacing $\;y$  with $\;-y$ :

$$13\mid 3x+2y\;\Leftrightarrow\;13\mid x+5y\;\Leftrightarrow\;13\mid 2x-3y\;\Leftrightarrow\;13\mid 5x-y. \tag{13a'}$$

The other relationships are:

$$13\mid  x-6y\;\Leftrightarrow\;13\mid 2x+y\;\Leftrightarrow\;13\mid 3x+5y\;\Leftrightarrow\;13\mid 5x-4y\;, \tag{13b}$$

$$13\mid x+6y\;\Leftrightarrow\:13\mid 2x-y\;\Leftrightarrow\;13\mid 3x-5y\;\Leftrightarrow\;13\mid 5x+4y\;, \tag{13b'}$$

$$13\mid x+2y\;\Leftrightarrow\;13\mid 4x-5y\;\Leftrightarrow\;13\mid 5x-3y\;\Leftrightarrow\;13\mid 6x-y\;, \tag{13c}$$

$$13\mid x-2y\;\Leftrightarrow\;13\mid 4x+5y\;\Leftrightarrow\;13\mid 5x+3y\;\Leftrightarrow\;13\mid 6x+y\;, \tag{13c'}$$

$13\mid x+3y\Leftrightarrow 13\mid 3x-4y\Leftrightarrow 13\mid 4x-y\Leftrightarrow 13\mid 5x+2y\Leftrightarrow 13\mid 6x+5y, \tag{13d}$

$13\mid x-3y\Leftrightarrow 13\mid 3x+4y\Leftrightarrow 13\mid4x+y\Leftrightarrow 13\mid 5x-2y\Leftrightarrow 13\mid 6x-5y, \tag{13d'}$

$13\mid x+4y\Leftrightarrow 13\mid 2x-5y\Leftrightarrow 13\mid 3x-y\Leftrightarrow 13\mid 4x+3y\Leftrightarrow 13\mid 5x-6y, \tag{13e}$

$13\mid x-4y\Leftrightarrow 13\mid 2x+5y\Leftrightarrow 13\mid 3x+y \Leftrightarrow 13 \mid 4x-3y\Leftrightarrow 13 \mid 5x+6y. \tag{13e'}$

          Let's demonstrate for example the equivalents from (13d).

$13\mid x+3y \Leftrightarrow \hat{1}\cdot x+\hat{3} \cdot y=\hat{0},$ where $\hat{m}$ mean the congruence class modulo 13  determined by $m$; their set is $\mathbb{Z}/13\mathbb{Z}$ (actually it is the field $GF(13)$. If we multiply last equation by $\hat{3},\;\hat{4},\;\hat{5},\;\hat{6}$ we obtain respectively

$$\hat{3}\cdot x +\hat{9}\cdot y=\hat{0} \Leftrightarrow \hat{3}\cdot x-\hat{4}\cdot y=\hat{0} \Leftrightarrow 13 \mid 3x-4y;$$

$$\hat{4} \cdot x+\hat{12} \cdot y=\hat{0} \Leftrightarrow \hat{4} \cdot x-\hat{1}\cdot y=\hat{0} \Leftrightarrow 13\mid 4x-y;$$

$$\hat{5}\cdot x+\hat{15} \cdot y=\hat{0} \Leftrightarrow \hat{5}\cdot x+\hat{2} \cdot y=\hat{0} \Leftrightarrow 13\mid 5x+2y;$$

$$\hat{6}\cdot x+\hat{18} \cdot y=\hat{0} \Leftrightarrow \hat{6} \cdot x+\hat{5} \cdot y=\hat{0} \Leftrightarrow 13\mid 6x+5y;$$

I used some elementary properties of $GF(13)$.

$\blacksquare$