My problem seems elementary. However a post in [SE](https://math.stackexchange.com/questions/4304278/least-common-multiple-of-three-integers) has not got an answer.


Let $a\_1,a\_2,a\_3\geq1$ be integers and let $A=\mathrm{lcm}(a\_1,a\_2,a\_3)$ be their least common multiple. I want to show the following.



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> If $m\_1,m\_2,m\_3\geq0$ are natural numbers satisfying $m\_1a\_1+m\_2a\_2+m\_3a\_3=2A$, then there exist integers $m\_1',m\_2',m\_3'$ satifying $0\leq m\_i'\leq m\_i$ for each $i$ and $m\_1'a\_1+m\_2'a\_2+m\_3'a\_3=A$.
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The motivation for this problem is the following problem I conjecture but can not prove either.



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> Let $\mathbb Q[x,y,z]$ be the polynomial ring of three variables. Let $a,b,c\geq1$ be the degrees of $x,y,z$ respectively. Let $A=\mathrm{lcm}(a,b,c)$. Then $\mathbb Q[x,y,z]\_{(A)}:=\bigoplus\_{d\geq0}\mathbb Q[x,y,z]\_{Ad}$ is generated by $\mathbb Q[x,y,z]\_A$. The subscript in $\mathbb Q[x,y,z]\_A$ denotes the degree $A$ subspace, i.e. $\mathbb Q[x,y,z]\_A=\mathrm{span}\big\{x^uy^vz^w:u,v,w\geq0,\;ua+vb+wc=A\big\}$.
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The first statement will prove the second. Consider $x^uy^vz^w\in\mathbb Q[x,y,z]\_{Ad}$ with $ua+vb+wc=Ad$.


* It suffices to show that there exist $\begin{cases}0\leq u'\leq u\\0\leq v'\leq v\\0\leq w'\leq w\end{cases}$ and $u'a+v'b+w'c=A$. This is because we then have $x^uy^vz^w=\big(x^{u'}y^{v'}z^{w'}\big)\big(x^{u-u'}y^{v-v'}z^{w-w'}\big)\in\mathbb Q[x,y,z]\_A\cdot\mathbb Q[x,y,z]\_{A(d-1)}$, and we reduce $d$ to $d-1$.
* Also if $ua\geq A$ (or $vb\geq A$ or $wc\geq A$), we just take $\begin{cases}u'=A/a\\v'=0\\w'=0\end{cases}$ as a solution.
* So the difficult situation is when $\begin{cases}0\leq u<A/a\\0\leq v<A/b\\0\leq w<A/c\end{cases}$. In this case $dA=ua+vb+wc<3A$, so $d<3$. This reduces to the $d=2$ case, which is the first statement.


Thanks for any comments.



Certain property of the least common multiple of three integers
1. Without loss of generality $d:=\gcd(a\_1,a\_2,a\_2)=1$, else divide $a\_1,a\_2,a\_3$ and $A$ by $d$.
2. Denote $d\_{ij}=\gcd(a\_i,a\_j)$. Now $d\_{12}, d\_{13},d\_{23}$ are mutually coprime, we may write $a\_1=d\_{12}d\_{13}b\_1$ etc, $A=d\_{12}d\_{13}d\_{23}b\_1b\_2b\_3$, $m\_1$ must be divisible by $d\_{23}$ etc, say $m\_1=n\_1d\_{23}$ and we get $n\_1b\_1+n\_2b\_2+n\_3b\_3=2b\_1b\_2b\_3$ for mutually coprime $b\_1$, $b\_2$, $b\_3$.
3. Assume the contrary. Denote $n\_1=k\_1b\_2+r\_1$ where $0\leqslant r\_1<b\_2$, $n\_2=k\_2b\_1+r\_2$ where $0\leqslant r\_2<b\_1$. Note that all numbers divisible by $b\_1b\_2$ and not exceeding $(k\_1+k\_2)b\_1b\_2$ are represented in the necessary form $t\_1b\_1+t\_2b\_2+t\_3b\_3$, $0\leqslant t\_i\leqslant n\_i$. Thus, assuming the contrary, we get $k\_1+k\_2\leqslant b\_3-1$, and $$n\_1b\_1+n\_2b\_2\leqslant (k\_1+k\_2)b\_1b\_2+(b\_2-1)b\_1+(b\_1-1)b\_2\\ \leqslant (b\_3-1)b\_1b\_2+(b\_2-1)b\_1+(b\_1-1)b\_2.$$
Summing up three such bounds we get $$4b\_1b\_2b\_3=2(n\_1b\_1+n\_2b\_2+n\_3b\_3)\leqslant 3b\_1b\_2b\_3+b\_1b\_2+b\_1b\_3+b\_2b\_3-2b\_1-2b\_2-2b\_3,$$
and $$(b\_1-1)(b\_2-1)(b\_3-1)<0,$$
a contradiction.


