$n$ divides $a_1 - a_2$ as well as $b_1 - b_2$. Show that $n$ divides $a_1b_1 - a_2b_2$.

Question

I keep arriving at $a_1b_1$ and $a_2b_2$ having the same sign if I multiply the equations $a_1 - a_2 = nk$ and $b_1 - b_2= np$ times each other. They must be opposite signs so that I can say that $n$ divides the difference of the two hence they are equivalent classes under $Z_8$. Any hints please. Thank you for your time.

This is the Congruence Product Rule in divisibility form. It is much clearer in congruence form. Follow the link for both. — Bill Dubuque, Apr 22 '15 at 15:58

score 5 · Accepted Answer · answered Apr 22 '15 at 15:48

5

$$ a_1 b_1 - a_2 b_2 = (a_1-a_2) b_1 + a_2(b_1-b_2). $$ $n$ divides both the brackets, so it also divides the sum.

answered Apr 22 '15 at 15:48

Chappers

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$n$ divides $a_1 - a_2$ as well as $b_1 - b_2$. Show that $n$ divides $a_1b_1 - a_2b_2$.

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