Is $$d_1\mid n,d_2\mid n\iff [d_1,d_2]\mid n$$ true ? And if yes, how can I prove it ? I recall that $[d_1,d_2]$ is the least common multiple.
My tries
For the implication : Let $d_1\mid n$ and $d_2\mid n$. Then, $d_1d_2\mid n^2$. By the way, since $d_1,d_2\mid n$, we have that $(d_1,d_2)\mid n$ where $(d_1,d_2)=:\gcd(d_1,d_2)$ and thus $$[d_1,d_2](d_1,d_2)\mid n^2.$$
Question: How can I get $[d_1,d_2]\mid n$ from this?
For the converse, since $d_1,d_2\mid [d_1,d_2]\mid n$, the claim follow.
Let $a$ and $b$ be integers such that $ad_1=bd_2=n$. By Bézout's theorem, there are integers $\alpha$ and $\beta$ such that $\alpha d_1+\beta d_2=(d_1,d_2)$. Therefore,$$(d_1,d_2)n=\alpha d_1n+\beta d_2n=\alpha d_1bd_2+\beta d_2ad_1=d_1d_2(\alpha b+\beta a).$$But this is the same thing as saying that $\frac{d_1d_2}{(d_1,d_2)}(\alpha b+\beta a)=n$. Since $\frac{d_1d_2}{(d_1,d_2)}=[d_1,d_2]$, this number divides $n$.
Theorem $\, a_1,\ldots,a_k\mid m\!\iff\! {\rm lcm}(a_1,\ldots,a_k)\mid m,\, $ for $\,a_i\neq 0\ \, $ [LCM Universal Property]
Proof $\,\ $ $(\Leftarrow)\,$ Note $\,{\rm lcm}(a_1,\ldots,a_k)\,$ is a common multiple of all $\,a_i\,$ by definition, thus $$\ a_i\mid {\rm lcm}(a_1,\ldots,a_k)\mid m\,\Rightarrow\, a_i\mid m\quad\text{by}\, {\it transitivity}\text{ of divisibility}$$
$(\Rightarrow)\ \ $ This has a natural conceptual proof by $\rm\color{#c00}{Euclidean\ descent}$ as follows.
The set $M$ of all positive common multiples of all $\,a_i\,$ is closed under $\rm\color{#0a0}{positive\ subtraction}$, i.e. $\,m> n\in M$ $\Rightarrow$ $\,a_i\mid m,n\,\Rightarrow\, a_i\mid m\!-\!n\,\Rightarrow\,\color{#0a0}{m\!-\!n}\in M.\,$ Thus $\,M\,$ is also closed under mod, i.e. remainder, by $\ m\ {\rm mod}\ n\, =\, m-qn = ((m-n)-n)-\cdots-n\,$ (= repeated subtraction). Thus the least $\,\ell\in M\,$ divides every $\,m\in M,\,$ else $\ 0\ne \color{#c00}{m\bmod \ell}\in M\,$ and is $\rm\color{#c00}{smaller}$ than $\,\ell.\,$ Note that $\,M\,$ is nonempty since $\,0\neq |a_1\cdots a_k|\in M$.
Remark $ $ The above means $\:\!\rm lcm\:\!$ is a divisibility-least common multiple, i.e. it is least in the divisibility order $\, a\!\prec b\!\!\! \overset{\rm def\!\!\!}\iff\! a\mid b.\ $ This is the (universal) definition of $\:\!\rm lcm\:\!$ used in general rings (which may lack any other way to measure "least"). See here for more on the general viewpoint. Note $\,0\mid m\iff m = 0,\,$ so any set containing $\:\!0\:\!$ has $\,\rm lcm = 0\,$ by this general definition.
Note that the proof used only that $M$ is a nonempty set of integers closed under subtraction, so it holds more generally for any such set of integers, i.e. every element is a multiple of the least positive element (if $M\neq \{0\}$).
That innate structure of the set of common multiples will be clarified when one studies ideal theory in abstract algebra, viz. a set of common multiples is a prototypical example of an ideal, and the above proof is a special case of: ideals in Euclidean domains (those enjoying Division with smaller Remainder) are necessarily principal - being the set of all multiples of any "least" element, i.e. Euclidean domains are PIDs. The proof is exactly the same as above by Euclidean descent using the division algorithm, i.e. directly using remainder as the descent step (which in more general rings cannot be obtained by repeated subtraction as in $\Bbb Z$). A prototypical example is the ideal of all polynomials that have some fixed algebraic number $\alpha$ as a root, which is princpal - generated by any minimal degree nonzero polynomial in the ideal - the so-called minimal polynomial of $\alpha$ (unique when normalized to be monic, i.e. lead coef $= 1)$.
Generalization $ $ follows by reformulating CRT in divisibility form as below.
Least Combined Multiple $ $ If $\,m,n\,$ are integers then
If $\,(m,n)\!=\!1\,$ then $\,\ \bbox[3px,border:1px solid #c00]{\begin{align} m&\mid M\\ n&\mid N\end{align}\!\!\iff mn\mid M + e(N\!-\!M)\,}\:\!,\,\ e = m\!\left[m^{-1}\bmod n\right]$
Similarly for the noncoprime case $\,\color{darkorange}{d := (m,n)} > 1\,$ we have
If $\,m,n\neq0\,$ then $\,\ \bbox[3px,border:1px solid #c00]{\begin{array}{} m\mid M\\ \ n\mid N\end{array}\!\!\iff\! \begin{align}\color{darkorange}{\gcd(m,n)}&\mid M,N \\ {\rm lcm}(m,n)&\mid M + e(N\!-\!M)\,\end{align}}\:\!,\,\ \begin{array}{} e = \bar m\!\left[\bar m^{-1}\bmod \bar n\right]\\ \:\!\bar m,\bar n =m/\color{darkorange}d,\, n/\color{darkorange}d\end{array}$
which follows by cancelling $\,\color{darkorange}d\,$ to reduce to the prior case for $\,m,n\,$ coprime.
Hint
If $n=p_1^{k_1}\cdot ...\cdot p_n^{k_n}$, then any divisor is of the form $$p_1^{\ell_1}\cdot ...\cdot p_n^{\ell_n},$$ with $0\leq \ell_i\leq k_i$. Ans if $d_1,d_2$ are two divisors, i.e.
$$d_1=p_1^{\ell_1^1}\cdot ...\cdot p_n^{\ell_n^1}\quad \text{and}\quad d_2=p_1^{\ell_1^2}\cdot ...\cdot p_n^{\ell_n^2} $$ then $$[d_1,d_2]=p_1^{\ell_1^1\wedge \ell_1^2}\cdot ...\cdot p_n^{\ell_n^1\wedge \ell_n^2},$$ where $a\wedge b$ mean the minimum of $a$ and $b$.
With this, the claim almost follow.
I think that the definition of LCM is as follows: m is said to be the lcm of $d_{1}$ and $d_{2}$ if
(1) $d_{1} |n, d_{2} |n$
(2) if $d_{1}|n$ and $d_{2}|n \implies m|n$
So, one side of the implication is just definition. And the reverse inclusion is also trivial. because $d_{1}|[d_{1} d_{2}]$ and so does $d_{2}$