Let any such
satisfying the desired result be called good
and let all other
be called bad
. I claim that
is good iff
First, we show
odd is bad.
Assume for the sake of contradiction that there is an odd number
that is good. Let
, where there are
one. Assuming that this does indeed become an n tuple of equal positive integers, let this
. Note that
Now, considering the reverse operation of the "sumop" or the "pomus"
we note that each pomus decreases both numbers. But, this will act on the numbers in pairs. As a result, one of the
's won't decrease.
We show that all
. Consider any
. Form the
. Observe that we can perform a "sumop" on
(what I mean is that we add
). So, we get all of the
's to have the same highest power of
dividing each of them (like
). Let this new
be the largest odd divisor of
be positive integers such that:
. We perform a "sumop" on
we note that we have a monovariant. So, since the
can't decrease infinitely, we find that eventually they are equal. Hence, we have all of them equal eventually.