TITLE: A question about LCM (number theory, supposedly high school level)
QUESTION [6 upvotes]: The question is: Determine whether or not there are positive integers $a, b, c$ such that: $\text{lcm}(a, b) = \text{lcm}(a + c, b + c)$.
I've tried writing $\text{lcm}(x,y)$ as $xy/\text{gcd}(x,y)$, but it didn't help whatsoever. How could one approach this problem using divisibility arguments and/or the F. T. of Arithmetic?

REPLY [2 votes]: Proof by contradiction:

assume $\text{lcm}(a,b)=\text{lcm}(a+c,b+c)=m$
$\exists(x\in\mathbb{Z},y\in\mathbb{Z}): [m=ax=by] \land [\text{gcd}(x,y)=1]$, ($x$ and $y$ are coprime)

if necessary, you can prove this by:
$\text{gcd}(x,y)=1$
otherwise $\exists(k):\frac{ki}{kj}=\frac{x}{y}$
... apply this to the above formula
and $\frac{m}{k}$ would be the "real" $\text{lcm}$

$\exists(x'\in\mathbb{Z},y'\in\mathbb{Z}): [m=(a+c)x'=(b+c)y'] \land [\text{gcd}(x',y')=1]$
$ax=m=(a+c)x'$, $by=m=(b+c)y'$
$ax=ax'+cx'$, $by=by'+cy'$
$a(x-x')=cx'$, $b(y-y')=cy'$
$a\frac{x-x'}{x'}=c$, $b\frac{y-y'}{y'}=c$
$c=a\frac{x-x'}{x'}=b\frac{y-y'}{y'}$, this looks like the original formula...
but $\frac{x-x'}{x'} \ne x$, $\frac{y-y'}{y'} \ne y$ if $x,y,x',y'$ are all integers
contradiction