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Over the domain of integers, if $\,a-c\mid ab+cd\,$ then $\,a-c\mid ad+bc$

Note: $x\mid y$ means "$x$ divides $y$," i.e. $\exists k\in \mathbb{Z}. y=x\cdot k$

This is part of an assignment on GCD, Euclidean algorithm, and modular arithmetic.

My approach:

If $a-c$ divides a linear combination of $a$ and $c$, then $a-c$ is a common divisor of $a$ and $c$. This comes from the definition of a common divisor: that if a certain $d$ divides two integers $x$ and $y$, then $d$ divides a linear combination of $x$ and $y$. Both $ab+cd$ and $ad+bc$ are linear combinations of $a$ and $c$ so $a-c$ must divide both of them.

Bill Dubuque
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VV6570
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  • The premise "$a-c$ divides a linear combination of $a$ and $c$" is always trivially true since $a-c$ itself is a linear combination of $a$ and $c$. Thus you cannot deduce anything from that statement alone. – Erick Wong Oct 26 '17 at 06:03

4 Answers4

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well, $ab + cd = ab - bc + bc + cd = b(a-c) + c(b+d)$ and thus:

$(a-c) | ab + cd \iff (a-c) | c(b+d)$

Likewise,

$ad + bc = ad - cd + cd + bc = d(a-c) + c(b+d)$.

So,

$(a-c) | ad + bc \iff (a-c) | c(b+d)$

Hence these conditions are equivalent:

That is, $(a-c) | ad + bc \iff (a-c) | ab + cd $.

Remark: My solution works, but provides little insight - i would love to see some geometric interpretations by others. e.g $(ab + cd) = (a,b) \cdot (b,c)$ - could we do something with this? and both $ad + bc$ and $ab + cd$ could be interpreted as the determinant of a matrix.

Maithreya Sitaraman
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  • This is part of a homework assignment on greatest common divisors, Euclidean algorithm and modular arithmetic...is there a way to prove it using these concepts? – VV6570 Oct 26 '17 at 03:34
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    perhaps. As a side note, i see you edited your question and entered an approach. Your approach is incorrect... because just because something divides a linear combination does not mean it divides individual terms. e.g 5| (2+3) but 5 neither divides 2 nor does it divide 3.. – Maithreya Sitaraman Oct 26 '17 at 04:03
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We know $$(a-c)|(a-c)$$ and $$(a-c)|(ab + cd) \tag{*}\label{*} $$ So we have : $$\begin{align}(a-c)|(a-c) &\Rightarrow (a-c)|(a-c)(b-d) \\ &\Rightarrow (a-c)|(ab -ad - bc +cd ) \\ & \Rightarrow (a-c) | (ad + bc - cd -ab) \\ &\stackrel{\eqref{*}}\Rightarrow (a-c) | (ad + bc)\end{align} $$

Q.E.D .

Shaun
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S.H.W
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Probably, some answers can be derived from each other. But, I also would like to leave a short answer of my own:

  1. Express $a-c$ with a single variable. Let $a-c=x$.

  2. Express $ad+bc$ in terms of $x$ as much as possible.

$$ad+bc=d(c+x)+b(a-x)$$

There is no need to open the parentheses. Just eliminate the terms containing $x$ and since $x\mid ab+cd\,$, you are done .

$\color{#a0a}{\rm{Second\,\,answer\,.}}$

Consider $a$ as a variable. Note that if $a-c\mid ab+cd $, then by Factor theorem setting $a=c$, we have :

$$cb+cd=0$$

Then, consider $c$ as a variable. Again by Factor theorem setting $c=a$, we have :

$$ab+ad=0$$

This implies that,

$$ad+bc=-(ab+cd)$$

which means $\,a-c\mid ad+bc\,$ .

  • If you leave a comment after the downvote, it will be easy to understand what went wrong with my answer. You may not upvote, however I did have a hard time understanding the quick downvote. I did not discover anything new. What I'm implementing is seeing what happens in divisibility by reducing the number of variables. There are more generalizations and more complicated implementations. –  Mar 05 '25 at 18:41
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    All $4$ of today's answers were downvoted, so the downvotes likely have nothing to do with math. – Bill Dubuque Mar 05 '25 at 19:44
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    It's instructive to compare answers [note $,f(x,y) = bx+dy,$ in OP] $$\begin{align} \text{me:}\ \ \bmod a!-!c!:\ a\equiv c\quad\ , \Rightarrow, &,f(a,c)\equiv f(c,a)\[.3em] \text{you:}\ \ \bmod\underbrace{a!-!c}_{\Large\color{#c00} \varepsilon}!:\ a\equiv c!+!\color{#c00}\varepsilon\Rightarrow, &,f(a,c) = f(c!+!\color{#c00}\varepsilon,a!-!\color{#c00}\varepsilon)\equiv f(c,a),,\ {\rm by},\ \color{#c00}{\varepsilon\equiv 0}\end{align}\qquad\quad\ \ \ $$ Mod arithmetic allows us to avoid introducing the extraneous null variable $,\color{#c00}{\varepsilon\equiv 0}.\ \ $ – Bill Dubuque Mar 05 '25 at 21:02
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    Re: 2nd proof in your update: again, it's essentially the same as the $,a,c,$ swap in my answer (cf. my prior comment) except you do the swap via an intermediate step $f(a,c)\equiv f(c,c)\equiv f(c,a),,$ and you use a roundabout way (Factor Theorem) to substitute congruent values, vs. simpler $,a\equiv c\Rightarrow f(a,c)\equiv f(c,c),$ etc. We don't usually derive the equation $,a=c\Rightarrow f(a,c) = f(c,c),$ in a roundabout way via Taylor / Factor Theorem, since it suffices to apply a simple substitution (i.e. equational logic). Ditto for congruences (generalized equations). – Bill Dubuque Mar 06 '25 at 03:35
  • @BillDubuque Thank you. Exactly. I almost understood your points . I would just ask this. If you were to check my exam paper, would you give full marks to the $2$ solutions I made ? Or, would you deduct points for certain reasons ? I mean, are there any missing or mathematical inconsistencies in my attempts ?Thank you. –  Mar 06 '25 at 14:16
  • To properly grade an exam requires much context that is lacking here (e.g. knowing what topics were taught, in what order, etc). Note: I have revised my answer to clarify the equational logic at the heart of the matter. $\ \ $ – Bill Dubuque Mar 09 '25 at 03:28
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It is worth explicitly highlighting how reformulating the divisibility relations into congruence equations makes the proof obvious - just swap congruent values $\,\color{#c00}{a\equiv c}$

$$\begin{align}&\!\bmod\, \overbrace{\color{#c00}{a\!-\!c}\!:}^{\large\color{#c00}{a\ \equiv\ c\ }}\ \ \ \overbrace{\color{#c00}ab\!+\!\color{#c00}cd}^{\color{#0a0} N}\,\equiv\, \overbrace{\color{#c00}cb\!+\!\color{#c00}ad}^{\color{darkorange}{N'}}\ \ \ (\text{by }{\bf swap} \,\ \color{#c00}{a,\:\!c})\\ &\text{therefore:} \ \ \color{#0a0}{N\equiv 0}\!\iff\! \color{darkorange}{N'\equiv 0^{\phantom{||^|}}}\\ &\text{hence:} \ \ \ \, a\!-\!c\mid \color{#0a0}N^{\phantom{|^|}}\!\!\!\! \iff\! a\!-\!c\mid \color{darkorange}{N'}\ \ \ \,\small\bf QED \end{align}\qquad\qquad$$

Generally we have that $\ a\!−\!c\mid \color{#0a0}{f(a,c)}\!\iff\! a−c\mid \color{darkorange}{f(c,a)}\,$ for any $\,f\in\Bbb Z[x,y],\,$ (i.e. for any polynomial $\,f(x,y)\,$ with integer coef's), simply by exploiting (swap) symmetry as we did above (and using the Polynomial Congruence Rule). OP is the special case $\,f(x,y)=bx+dy.\,$

Key idea converting the divisibility to congruence form enables us to make deductions using well-known (mod) arithmetic and equational logic, i.e. OP boils down to the congruence form of the equational inference $\,a=c\,\Rightarrow\, f(a,c) = f(c,a).\,$ One of the primary motivations for congruence language is that it makes it simple to reuse our well-honed intuition on manipulation of equations, e.g. substituting equal values, peforming the same operation on both sides, etc. So it is essential to learn to think of congruences as generalized equations.

Our inference above: $\,m\mid \color{#0a0}{N}\!\iff\! m\mid \color{darkorange}{N'}\ $ if $\ \color{#0a0}{ N\!\equiv\! 0}\!\iff\! \color{darkorange}{N'\!\equiv\! 0}\pmod{\!m}\,$ is a special case of ubiquitous divisibility mod reduction, which one should know to be proficient at mod arithmetic. $\ \ $

Bill Dubuque
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  • Because $a \equiv a-(a-c) = c\pmod {a-c}$ – JMP Mar 05 '25 at 17:25
  • @JMP $,a-c\equiv 0,\Rightarrow, a\equiv c,$ by adding $,c,$ to both sides, same as for $\rm\color{#0a0}{equations}$. The reason we use congruences is that they reduce non-intuitive $\rm\color{#c00}{divisibility}$ arithmetic to familiar $\rm\color{#0a0}{equation}$ arithmetic.$\ \ $ – Bill Dubuque Mar 05 '25 at 17:53