I have the following question. Is it possible to find such natural numbers $a$, $b$, $c$ such that
$lcm(a,b)+lcm(b,c)=lcm(a,c)$
where $lcm$ is the least common multiple?
My thoughts about this:I know that $lcm(a,b)=\prod_{i=1}^{i=n}p_i^{\max(a_i,b_i)}$ and $lcm(b,c)=\prod_{i=1}^{i=n}p_i^{\max(b_i,c_i)}$ and $lcm(a,c)=\prod_{i=1}^{i=n}p_i^{\max(a_i,c_i)}$. I don't know how to find answer on this question from this. Thank you!
The equation being investigated is
$$\operatorname{lcm}(a,b) + \operatorname{lcm}(b,c) = \operatorname{lcm}(a,c) \tag{1}\label{eq1A}$$
The term "natural numbers", for some people, includes $0$. However, note that, as stated in least common multiple, the $\operatorname{lcm}$ function is usually not defined if any argument is $0$. Nonetheless, consider defining $\operatorname{lcm}(n,0) = n$, as some authors do. Then, if $a = 0$, \eqref{eq1A} becomes $b + \operatorname{lcm}(b,c) = c \;\to\; \operatorname{lcm}(b,c) = c - b$. Since $\operatorname{lcm}(b,c) \ge \max(b,c)$, this means that $b = 0$, with $c$ being any value.
If $b = 0$, we get $a + c = \operatorname{lcm}(a,c)$. With $a = c$, we have $2a = a$, so $a = c = 0$. If $c \gt a$, then note $a = \operatorname{lcm}(a,c) - c$. Since $c \mid \operatorname{lcm}(a,c)$, then $c \mid a$, which is only possible if $a = 0$. Similarly, we have that if $a \gt c$, then $c = 0$.
Finally, if $c = 0$, then \eqref{eq1A} gives $\operatorname{lcm}(a,b) = a - b$ so, as explained earlier, we have $b = 0$ and $a$ can be any value.
Otherwise, with $a$, $b$ and $c$ all being positive integers, there are no solutions. To show this, using the $p$-adic valuation function, consider the relative sizes of
$$\nu_2(a) = a_2, \;\; \nu_2(b) = b_2, \;\; \nu_2(c) = c_2 \tag{2}\label{eq2A}$$
If there's one largest value, say $a_2 \gt b_2$ and $a_2 \gt c_2$, then we have $\nu_2(\operatorname{lcm}(a,c) - \operatorname{lcm}(a,b)) \ge a_2 + 1$ (since $\operatorname{lcm}(a,c) - \operatorname{lcm}(a,b) = 2^{a_2}d_1 - 2^{a_2}d_2 = 2^{a_2}(d_1 - d_2)$ where $d_1$ and $d_2$ are odd, so $d_1 - d_2$ is even) but $\nu_2(\operatorname{lcm}(b,c)) \lt a_2$, contradicting \eqref{eq1A}. Similarly, we can't have just $b_2$ or $c_2$ being the largest.
Instead, if there's two largest values, say $a_2 = b_2 \gt c_2$, then we get $\nu_2(\operatorname{lcm}(a,b)) = a_2$ while $\nu_2(\operatorname{lcm}(a,c) - \nu_2(\operatorname{lcm}(b,c)) \ge a_2 + 1$, which contradicts \eqref{eq1A}. Similarly, we can't have just $a_2$ or $b_2$ being the smallest.
Finally, if there are three largest values, i.e., they are all the same, so $a_2 = b_2 = c_2$, then $\nu_2(\operatorname{lcm}(a,b) + \operatorname{lcm}(b,c)) \ge a_2 + 1$ but $\nu_2(\operatorname{lcm}(a,c)) = a_2$, once again contradicting \eqref{eq1A}.
Since this covers all possible cases, we get that \eqref{eq1A} is never true for any positive integers $a$, $b$ and $c$.