Let be natural numbers.

Does

, where is number of positive divisors of k, implies that

?

Thanks.

Printable View

- June 1st 2010, 07:18 PMmeleseIs the converse statement true?
Let be natural numbers.

Does

, where is number of positive divisors of k, implies that

?

Thanks. - June 1st 2010, 07:27 PMchiph588@
- June 1st 2010, 07:29 PMBruno J.
You probably mean ?

- June 1st 2010, 07:30 PMBruno J.
- June 1st 2010, 07:46 PMchiph588@
- June 1st 2010, 09:27 PMBruno J.
If you prove it, then it's true! (Happy)

Here's a less computational argument. Let denote the set of divisors of , and suppose that . I exhibit a surjection which is not an injection. For every , let . This map is clearly a surjection. But . Therefore . - June 2nd 2010, 07:33 AMmelese
yes I do, I wrote that early in the morning...

- June 2nd 2010, 07:44 AMmeleseI have an argument but I'm not sure...
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. - June 2nd 2010, 11:50 AMchiph588@
- June 3rd 2010, 06:23 AMmelese
Again, you corrected me. Thank you so much for your help!