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.