# Divisibility problem: show $(x-z)\mid xy+zw \implies (x-z)\mid xw+yz$

I'm stuck at this homework problem can someone help me out? Much appreciated!

$$(x-z)\mid xy+zw \implies (x-z)\mid xw+yz$$ Thanks again!

• Can you share what you've done so we can better assist? Mar 18 '13 at 14:47

## 3 Answers

It's actually pretty simple:

$$(xy+wz)-(xw+yz)=x(-w+y)-z(-w+y)=(x-z)(-w+y)$$

$\begin{eqnarray}\rm {\bf Hint}\ \ mod\ x\!-\!z\!:\,\ \color{#C00}x\equiv \color{#0A0}z\ &\Rightarrow&\rm\, \ \ \ \ \ \ \color{#C00}x\,y+\color{#0A0}z\,w\\ &&\rm\ \equiv\, \color{#0A0}z\,y + \color{#C00}x\,w\end{eqnarray}$

Remark $\$ Notice how much simpler it is using congruences (equational reasoning) vs. divisibility (relational relasoning). Using congruences the proof is obvious: simply replace equals by equals. This simplification works quite generally. It allows us to convert less familiar relational statements to equivalent equational statements - where we can employ our well-practiced manipulation of equations (viewing congruences as generalized equations).

• Coloring the fonts must be very difficult.? I like the colored parts ;). +1 for the coloring. Mar 18 '13 at 15:17

Hint: Expand the polynomial $(x-z)(y-w)$.