Let $m, \ n, \ k \in \Bbb N $ be such that $ \operatorname{lcm}[m , m + k] = \operatorname{lcm}[n , n + k],$ then prove that $ m = n.$
Though I wasn't able to proceed much, but here is a sketch of what I tried.
First let $l = \operatorname{lcm}(m, m + k) = \operatorname{lcm}lcm(n, n + k)$
now we have $ m \mid l, \ (m + k) \mid l, \ n \mid l, \ (n + k)\mid l $ then
next since $\gcd(m , m + k) \cdot \operatorname{lcm}(m, m + k) = m \cdot (m+k)$
but we also have $\gcd(m, m + k) = \gcd(m, k)$ and now we have $ \frac{m \cdot (m + k)}{\gcd(m, k)} = \frac{n \cdot (n + k)}{\gcd(n, k)} $.
Now I noticed if $ m \mid k $, then we are done, but that is not always possible, so I let $\gcd(m, k) = d_{1}$ and $\gcd(n, k) = d_{2}$ and started replacing, but it becomes more and more difficult in that way, so I quit here,
Then I also thought to consider it as an equation asking us to solve $\operatorname{lcm}(m, m + k) = l$ but again since after dividing by $\gcd(m, k)$ it will give two corresponding quadratic so, this method also failed.
Also, the book I am using hasn't introduced much of congruence and even the fact that $\gcd(a, b) \cdot \operatorname{lcm}(a,b) = a \cdot b $ so a solution without that will be nicer.
Any help/hints are appreciated, Thanks in advance.
Best Answer
$\textbf{Hint:}$ Consider one prime at a time.Say we have a prime $p$.
Its highest exponent dividing $m,n,k$ are respectively $a_1,a_2,a_3$ Now,if we can deduce that $a_1=a_2$ for every prime we consider,we would be done.
$\textbf{Solution:}$consider one side: $lcm(m,m+k)$. If $a_1 \ge a_3$ then, $m+k$ has highest exponent $a_3$ and m has $a_1$.then,$lcm(m,m+k)$ has $a_1$ since its the greater one.
Again,if $a_1 \le a_3$ then,$m+k$ has highest exponent $a_1$ ,so is $m$. Hence, $lcm(m,m+k)$ has $a_1$ as its highest divisor. same is true for the other side of the equation.
Since,both sides are equal $a_1=a_2$ for every prime we consider.