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$a,b\in\mathbb Z^+$, 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?

Bill Dubuque
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Aiden Chow
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    If you use \text{lcm} rather than \operatorname{lcm} then when you type 3\text{lcm}(a,b) you'll see $3\text{lcm}(a,b)$ instead of $3\operatorname{lcm}(a,b).$ And it's not just that some horizontal space is added; rather the amount of space depends on the context. $$\begin{align} & 3\operatorname{lcm}(a,b) \ & 3\text{lcm}(a,b) \end{align} $$ – Michael Hardy Sep 14 '20 at 18:26
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    As in the dupe, cancelling $\gcd(a,b)$ reduces to the coprime case, where we need only find factors of $105$ satisfying the other condition (here: their difference divides $120$), and there are only finitely many (few) possible divisors to check. As explained there, this homogeneous reduction is a commonly used technique to simplify such problems so you should be sure to master it (see links there for more examples) – Bill Dubuque Sep 14 '20 at 18:57

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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.

Michael Rozenberg
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