relatively prime integers....

• Mar 16th 2008, 09:30 AM
pizzaRusher1234
relatively prime integers....
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.
• Mar 16th 2008, 11:58 AM
Aryth
If a is divisible by both b and c, we have:

\$\displaystyle bk = a\$

\$\displaystyle cj = a\$

\$\displaystyle b = a/k\$

\$\displaystyle c = a/j\$

Therefore:

\$\displaystyle bc = a/jk\$

\$\displaystyle jk = i\$

\$\displaystyle bc(i) = a\$

Thus:

\$\displaystyle bc | a\$
• Mar 16th 2008, 12:37 PM
pizzaRusher1234
Quote:

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

\$\displaystyle bk = a\$

\$\displaystyle cj = a\$

\$\displaystyle b = a/k\$

\$\displaystyle c = a/j\$

Therefore:

***\$\displaystyle bc = a/jk\$***

\$\displaystyle jk = i\$

\$\displaystyle bc(i) = a\$

Thus:

\$\displaystyle bc | a\$

ummm....\$\displaystyle bc = a^2/jk\$. how does that work?
• Mar 16th 2008, 01:00 PM
JaneBennet
b and c divide a \$\displaystyle \Rightarrow\$ \$\displaystyle a=hb=kc\$ for some integers h and k.

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

Now multiply through by a.

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