Is the equality $(x+y) \cdot lcm(x,y) = y \cdot lcm(x,x+y)$ true?

Question

Is the equation I wrote in the title true for positive integers $x,y$? I checked some cases and it seems to hold, but how do I prove it? I am trying to solve another problem and it turns out that if this equation holds, the problem is solved.

Answer 1

It is true:

$$\begin{array}{rcl}(x+y)\cdot \text{lcm}(x,y)&=&(x+y)\cfrac{xy}{\gcd(x,y)}\\&=&y\cfrac{x(x+y)}{\gcd(x,x+y)}\\&=&y\cdot \text{lcm}(x,x+y)\end{array}$$

I've used two facts here:

$$\text{lcm}(x,y)\cdot\gcd(x,y)=xy$$

(Consider the powers of every prime going into $\text{lcm}(x,y)$, $\gcd(x,y)$, $x$ and $y$.)

$$\gcd(x,y)=\gcd(x,x+y)$$

(Follows because every divisor of $x$ and $y$ is a divisor of $x$ and $x+y$ and vice versa.)

Answer 2

Yes, it's true. Recall that $$\text{lcm}(x,y)={xy\over\gcd(x,y)}$$ Since $\gcd(x,y)=\gcd(x,x+y)$, a simple substitution proves your formula.

Answer 3

Hint: $\\lcm(x,y)=\frac{xy}{\gcd(x,y)}$ and $\gcd(x,x+y)=\gcd(x,y)$.

Answer 4

It can be proved directly by lcm laws (without using gcds) as follows, with $\,[a,b] := {\rm lcm}(a,b)$

$$\begin{align} \color{#c00}{\dfrac{n}{yz}}\:\ +\:\ \dfrac{n}{xz}\:\ =\:\ \dfrac{n}{xy},\,\ \ z = x\!+\!y \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\[.5em] \Rightarrow\ \ {\rm if}\ \,\color{#c00}{yz\mid n}\,\ {\rm then}\,\ xz\mid n\!\iff\! xy\mid n\\[.5em] \Rightarrow\ \ xz,\color{#c00}{yz}\mid n\!\iff\! xy,\color{#c00}{yz}\mid n\\[.5em] \Rightarrow\ \ \underbrace{[xz,yz]}_{\textstyle z\,[x,y]}\ =\ \underbrace{[xy,yz]}_{\textstyle y\,[x,z]}\ \ \ \ \\ \end{align}\qquad\qquad\qquad\qquad$$

The final line applies the LCM Universal Property and the LCM Distributive Law.