This problem has been originally posted at [math.stackexchange](https://math.stackexchange.com/q/4746506/573047).


Since there are no answers and no comments there yet, I am crossposting it here to know if it is already known and tractable.


Here is the problem, slightly restricted with respect to the original:


starting from a multiset of $6$ integer numbers, we modify it repeatedly by replacing two elements $a,b$ at choice with $a+b,a+b$. Is it possible to make all elements equal for any choice of the original multiset?
Does the answer change if we replace "integer" with "rational"?


Note that the original poster at math.stackexchange seems to have a proof for $n=3,5$ elements.
As noted in comments below, it is impossible to equalize $a,a,b$ with $a \gt b \ge 0$, because you will have another sequence of the same form.



"Make all numbers equal" game
I claim that it is always possible to make all of the integers equal to each other. To see this, consider the allowable transformations


1. $(a,b,c,d,e,f)\mapsto(a+b,a+b,c,d,e,f)$,
2. $(a\_1,\dots,a\_6)\mapsto(a\_{\sigma(1)},\dots,a\_{\sigma(6)})$, and
3. $(a,b,c,d,e,f)\mapsto(a/n,b/n,c/n,d/n,e/n,f/n)$ whenever $a/n,b/n,c/n,d/n,e/n,f/n$ are all integers.


We shall write $(a,b,c,d,e,f)\mapsto^\*(g,h,i,j,k,l)$ if $(g,h,i,j,k,l)$ can be obtained from $(a,b,c,d,e,f)$ by using a sequence of allowable transformations. We observe that if $(a,b,c,d,e,f)\mapsto^\*(1,1,1,1,1,1)$, then can transform $(a,b,c,d,e,f)$ into $(n,n,n,n,n,n)$ for some $n$ using only transformations $1$ and $2$, but transformation 3 makes it easier to use induction to show that we can always get $(a,b,c,d,e,f)\mapsto(1,1,1,1,1,1)$.


We observe that $(a,b,c,d,e,f)\mapsto^\*(a+b+c+d,a+b+c+d,a+b+c+d,a+b+c+d,e+f,e+f)$.
Now set, $\alpha=a+b+c+d,\beta=e+f$. I claim that $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)=(0,0,0,0,0,0)$ or $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^\*(1,1,1,1,1,1)$ by induction on the square of the $\ell^2$ norm $4\alpha^2+2\beta^2$ of $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)$.


Case 0: $\alpha=0,\beta=0$. This is trivial.


Case 1: $\alpha=0,\beta\neq 0$. In this case,
$(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^\*(\beta,\beta,\beta,\beta,\beta,\beta)\mapsto^\*(1,1,1,1,1,1)$.


Case 2: $\alpha\neq 0,\beta=0.$


$(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^\*(\alpha,\alpha,\alpha,\alpha,\alpha,\alpha)\mapsto^\*(1,1,1,1,1,1)$.


Case 3: $\alpha\neq 0,\beta\neq 0$, $\alpha$ is even.


If $\alpha$ is even, then
$(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^\*(\alpha,\alpha,\alpha,\alpha,2\beta,2\beta)\mapsto^\*(\alpha/2,\alpha/2,\alpha/2,\alpha/2,\beta,\beta)$ and
$(\alpha/2,\alpha/2,\alpha/2,\alpha/2,\beta,\beta)\mapsto^\*(1,1,1,1,1,1)$ by the induction hypothesis.


Case 4: $\alpha\neq 0,\beta\neq 0$, $\beta$ is even.
$(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^\*(2\alpha,2\alpha,2\alpha,2\alpha,\beta,\beta)\mapsto^\*(\alpha,\alpha,\alpha,\alpha,\beta/2,\beta/2)$, and
$(\alpha,\alpha,\alpha,\alpha,\beta/2,\beta/2)\mapsto^\*(1,1,1,1,1,1)$ by the induction hypothesis.


Case 5: $\alpha\neq 0,\beta\neq 0$, $\alpha,\beta$ are both odd.


Let $\alpha=2\gamma+1,\beta=2\delta+1.$ Then $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)
\mapsto^\*(2(2\gamma+1),2(2\gamma+1),2(\gamma+\delta+1),2(\gamma+\delta+1),2(\gamma+\delta+1),2(\gamma+\delta+1))$
$\mapsto^\*(2\gamma+1,2\gamma+1,\gamma+\delta+1,\gamma+\delta+1,\gamma+\delta+1,\gamma+\delta+1)=(\alpha,\alpha,\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2}).$


If $\alpha=\beta$, then clearly $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)
\mapsto^\*(1,1,1,1,1,1)$, and if $\alpha\neq\beta$, then we may use the induction hypothesis to conclude that $(\alpha,\alpha,\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2})\mapsto^\*(1,1,1,1,1,1).$


