If $(a-c)|(ab+cd)$ then $(a-c)|(ad+bc)$.

Question

Over the domain of integers, if $(a-c)|(ab+cd)$ then $(a-c)|(ad+bc)$.

Note: $x|y$ 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.

Answer 1

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 $.

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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.

Answer 2

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 .