If $b_1\mid a$ and $b_2\mid a$ and $\gcd(b_1,b_2)=1,$ show that $b_1b_2\mid a$.

Question

I am trying to prove the following statement: if $b_1\mid a$ and $b_2\mid a$ and $\gcd(b_1,b_2)=1,$ show that $b_1b_2\mid a$.

Well for starters we can write $a=b_1q_1$ and $a=b_2q_2$ for some $q_1,q_2\in\mathbb{Z}$.

I am not sure how I am supposed to use the fact that $\gcd(b_1,b_2)=1$.

We have $a=b_1q_1=b_2q_2$, so I think we must have $q_1=kq_2$ for some $k\in\mathbb{Z}$, as otherwise they cannot be equal.

Answer 1

This statement follows pretty easily from the uniqueness of the prime decomposition. The prime powers that divide $b_1$ and the prime powers that divide $b_2$ (the two sets of primes are disjoint, since $b_1, b_2$ are coprime) must all appear in the decomposition of $a$, and so we get that the product of all these prime powers, which is $b_1b_2$, divides $a$.

However, this can also be proved without using prime decompositions. Since $\gcd(b_1,b_2)=1$, there are some $k,l\in\mathbb{Z}$ such that $kb_1+lb_2=1$. Now multiply this equality by $a$ and see what you get. Hint: Remember that instead of $a$ you can substitute $q_1b_1$ or $q_2b_2$ if needed.