A question about LCM (number theory, supposedly high school level) 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?
 A: 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

A: First we assert that $a \nmid b$, for if otherwise then $\text{lcm}(a,b)=b <\text{lcm}(a+c,b+c)$ for any positive $c$. Then it follows that $b \nmid a$ by symmetry.  
We define $k$ so that $k=\text{lcm}(a,b)$, so there exist some $e_1$ and $e_2$ so that $a \cdot e_1 = k$ and $b \cdot e_2 = k$. Assume there exists a $c$ so that $\text{lcm}(a+c,b+c)=k$. Then there exist some $d_1$ and $d_2$ so that $(a + c) \cdot d_1 = k$ and $(b+c) \cdot d_2 = k$ so $(a \cdot d_1) + (c \cdot d_1) = k$ (* ) and $(b \cdot d_2) + (c \cdot d_2) = k$ (** ). Now clearly $d_1<e_1$ and $d_2<e_2$ (because $c \cdot d_1 > 0$ and $c \cdot d_2 > 0$) so we get $d_1 | e_1$ and $d_2 | e_2$ (because $k$ can only be factored in one way; here is where the FTA comes in), so $d_1 | (k/a)$ and $d_2 | (k/b) $ so we can define new numbers $m_1$ and $m_2$ so that $d_1 \cdot m_1 = k/a$ and $d_2 \cdot m_2 = k/b$. Plugging this into (* ) and (** ) gives us $k / m_1 + c \cdot k / (m_1 \cdot a) = k \iff 1 / m_1 + c / (m_1 \cdot a) = 1$ and analogously we get $1 / m_2 + c / (m_2 \cdot b) = 1$, which results in the following equations:
i) $b+c=m_2 \cdot b$
ii) $a+c=m_1 \cdot a$. 
Equation ii) gives us that $c=(m_1-1)a$ so by plugging that into equation i), we get $b+(m_1-1)a = m_1 * b \iff a/b=(m_2-1)/(m_1-1)$ which contradicts that $b \nmid a$. Thus our assumption about $c$ was false so we have proven by contradiction that the integers $a,b,c$ do not exist as described. Q.e.d.
