$a,b\in\mathbb Z^+$, and the equations given are $a-b=120$, and $\operatorname{lcm}(a,b)=105\gcd(a,b)$. What is $a$?

Question

$a,b\in\mathbb Z^+$, and the equations given are $a-b=120$, and $\operatorname{lcm}(a,b)=105\gcd(a,b)$. What is $a$?

So what I did was that I first found that $a=b+120$, and plugged that value of $a$ into the second equation:$$\begin{align}\operatorname{lcm}(120+b,b)&=105\gcd(120+b,b)\\\frac{ab}{\gcd(120+b,b)}&=105\gcd(120+b,b)\\\gcd(120+b,b)^2&=\frac{ab}{105}\\\gcd(120+b,b)&=\sqrt{\frac{ab}{105}}\\\gcd(120,b)&=\sqrt{\frac{b(b+120)}{105}}\end{align}$$I'm not too sure if this is the correct way to go. Should I manipulate the equations, or should I do something else?

Answer 1

Let $a=dx$ and $b=dy,$ where $d=\gcd(a,b).$

Thus, $$ab=105d^2$$ or $$xy=105=1\cdot3\cdot5\cdot7.$$ Also, $$d(x-y)=120,$$ which says that $120$ is divisible by $x-y$.

Can you end it now?

I got $(x,y)=(15,7)$ is valid.