relatively prime integers....

**pizzaRusher1234** · March 16th 2008, 09:30 AM

Hello everbody...

Let b and c be relatively prime integers, and suppose a is an integer that is
divisible by both b and c. Prove that bc|a.

**Aryth** · March 16th 2008, 11:58 AM

If a is divisible by both b and c, we have:

$bk = a$

$cj = a$

$b = a/k$

$c = a/j$

Therefore:

$bc = a/jk$

$jk = i$

$bc(i) = a$

Thus:

$bc | a$

**pizzaRusher1234** · March 16th 2008, 12:37 PM

Originally Posted by Aryth

If a is divisible by both b and c, we have:

$bk = a$

$cj = a$

$b = a/k$

$c = a/j$

Therefore:

*** $bc = a/jk$ ***

$jk = i$

$bc(i) = a$

Thus:

$bc | a$

ummm.... $bc = a^2/jk$ . how does that work?

**JaneBennet** · March 16th 2008, 01:00 PM

b and c divide a $\Rightarrow$ $a=hb=kc$ for some integers h and k.

b and c are relatively prime means there are integers $p,q$ such that $pb+qc=1$ .

Now multiply through by a.

$pba+qca=a$ $\Rightarrow$ $a=pb(kc)+qc(hb)=(pk+qh)bc$ .

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