Allowing the exponents to be zero, if necessary, we can write the following (similar) prime factorizations:
and
, where for
and
are nonnegative. Also, for any prime
,
divides
at leat one of and
.
Now,
and then
.
Computing
gives
.
Now comes my difficulty...
For each
,
, so if we want
we must have
for
. Otherwise
.
But then
. Without loss of generality
and this means that
divides n but not m. In this manner
and
are relatively prime.
By the way exactly one of
and
equals
due to: "for any prime
,
divides
at leat one of and
."
Is my argument right? Thanks for your help.