If $p|(a-b)$ and $q|(a-b)$, then prove that $pq|(a-b)$

Question

I'm quite familiar with the definition of multiple(s) but I can't seem to be able to prove that if $p|(a-b)$ and $q|(a-b)$, then $pq|(a-b)$. Also, $p$ and $q$ are relatively prime.

Answer 1

You want to prove that if $p,q$ are coprime, and $p|n$, $q|n$, then $pq|n$.
Write $n=k_1p=k_2q$, then since $p|k_2q$ and $(p,q)=1$, it must happen $p|k_2$ and then $pq|k_2q=n$.
The assumption that $p$, $q$ are coprime is not removable, for you can take $p=4$, $q=6$, then both $p$, $q$ divide $12$, but $pq=24$.

Answer 2

For simplicity, let $a-b = c$. Then the statement becomes, if $p$ and $q$ are relatively prime and $p|c$ and $q|c$, then $pq|c$.

PROOF: By Fundamental Theorem of Arithmetic, we can decompose $p$ and $q$ as $p = p_1^{r_1}\cdot p_2^{r_2}\cdot...\cdot p_k^{r_k}$ and $q = q_1^{s_1}\cdot q_2^{s_2}\cdot...\cdot q_l^{s_l}$ where $p_1,...,p_k, q_1,...,q_l$ are prime numbers and $r_1,...,r_k,s_1,...,s_r \ge 1$. Now, if $p$ and $q$ are relatively prime, then there is no mutual prime divisor of $p$ and $q$, i.e., $p_1,...,p_k, q_1,...,q_s$ are all distinct. And since both $p$ and $q$ divide $c$, we can say $c = p_1^{r_1}\cdot p_2^{r_2}\cdot...\cdot p_k^{r_k}\cdot q_1^{s_1}\cdot q_2^{s_2}\cdot...\cdot q_l^{s_l}\cdot k$ for some integer $k$ and we are done.

Answer 3

One way to proved that is to check the prime factorization of $(a-b)$

Note that every prime which appears in $p$ or $q$ it also appears in $a-b$ with at least the same power.

Since $p$ and $q$ are relatively prime they do not share any prime, so when you multiply them all the primes in the product appears in $a-b$ with the same or higher power.