$\text{lcm} (a, b)=\text{lcm} (a+c, b+c)$ 
Can $\text{lcm} (a, b)=\text{lcm} (a+c, b+c)$ for natural $a, b, c$?

I've tried writing out all divisors of $a, b, c$ like $p_1 p_2$ etc. And tried that maybe if $a+c> a$ and $b+c> b$ the $\text{lcm}$ must be greater. 
 A: This is not true in general as we see
lcm$(3,4)=12$ whereas lcm$(3+3,4+3)=42$
One more note, as you stated that $a,b,c$ are all naturals hence $c>0$ and thus always $a+c>a,b+c>b$ But you see that lcm$(7,9)=63$ whereas lcm$(7+1,9+1)=40<63$
A: Actually, lcm$(a,b)$ and lcm$(a+c,b+c)$ are never equal.
Why? The point is to investigate the prime factors of the numbers, and look at the minimal triplet satisfying the condition. This leads to impossibility.
Assume otherwise: there would triplet $(a_{0},b_{0},c_{0})$ for which the condition is satisfied, with minimal product $a_{0}b_{0}c_{0}$. If $\gcd(a_{0},b_{0}) > 1$, we could write $a_{0} = p a_{0}'$ and $b_{0} = p b_{0}'$ for some prime $p$. Now $p$ | lcm$(a_{0},b_{0}) $ so $p$ | lcm$(a_{0} + c_{0},b_{0}+c_{0}) $. Now $p$ would have to divide at least one of the $a_{0} + c_{0}$ and $b_{0}+c_{0}$, both lead to $p$ | $c_{0}$, so $c_{0} = p c_{0}'$. Now since lcm$(k a,k b)$ = $k$ lcm$(a,b)$ we see that also $(a_{0}',b_{0}',c_{0}')$, satisfies the condition, contradicting the minimality.
Now $\gcd(a_{0},b_{0}) = 1$, so lcm$(a_{0},b_{0}) = a_{0}b_{0}$. Assuma now that $\gcd(a_{0}+c_{0},b_{0}+c_{0}) > 1$. Now some prime $p$ divides $a_{0}+c_{0}$ and $b_{0}+c_{0}$, so also their difference $a_{0} -b_{0}$. The prime would also divide right-hand side of the equality, lcm$(a+c,b+c)$, so also $ab$. If $p$ | $a_{0}$, then also $p$ | $b_{0}$, as $b_{0} = a_{0} - (a_{0} -b_{0})$. But now $\gcd(a_{0},b_{0}) > 1$, contradiction. So $\gcd(a_{0}+c_{0},b_{0}+c_{0}) = 1$ and therefore lcm$(a_{0}+c_{0},b_{0}+c_{0}) = (a_{0}+c_{0})(b_{0}+c_{0})$.
Finally, since $(a_{0}+c_{0})(b_{0}+c_{0}) > a_{0}b_{0}$ for $c_{0} > 0$, we can't have equality between lcm$(a_{0},b_{0})$ and lcm$(a_{0}+c_{0},b_{0}+c_{0})$, and we are done.
