let n be an integer greater than 1. Pick 2 distinct divisors of n. What is the probability that EXACTLY one of the divisors is a perfect square? This has been approved by Captain Black.

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- May 24th 2011, 10:20 AMChris11Divisibility Probability
let n be an integer greater than 1. Pick 2 distinct divisors of n. What is the probability that EXACTLY one of the divisors is a perfect square? This has been approved by Captain Black.

- May 24th 2011, 08:07 PMchisigma
In order to undestand correctly: what does it mean 'exactly one of the divisor is a perfect square?'...

May be...

$\displaystyle 6=3*2$ has no square factors...

$\displaystyle 12=3*2^{2}$ has*exactly*one square factor...

$\displaystyle 36=3^{2}*2^{2}$ has more than one square factor...

... so that 12 is a 'candidate', 6 and 36 aren't?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - May 24th 2011, 10:18 PMChris11
Hey. By example. Consider 36.

Pick 36, 3. 3 isn't a perfect square, but 36 is. That's what I mean by 'exactly one of the divisors is a perfect square.' - May 25th 2011, 02:51 AMtonio

Hmmm...I think 1 isa perfect square divisor of any natural...__always__

But even then there are some doubts:

1) Is the integer number n chosen randomly or it is given and THEN we begin choosing randomly from its divisors?

2) Are we supposed to assume that when we choose randomly (I guess) two divisors then

these divisors are different or is it possible that both are one and the same?

Tonio - May 25th 2011, 10:49 AMChris11
1. You're suppose to find an expression for all n.

2. You pick any 2 distinct divisors at random. You pick one, then your universe becomes the set of divisors that you haven't picked. Then you pick one from this set.

3. 1 is something that you 'pick' in this case. Please don't think that you pick 1 automaticly when you pick 57. - May 26th 2011, 10:56 PMChris11
Warning There Is a hint below

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Consider the prime power factorization of n. - May 30th 2011, 06:07 AMSpringFan25
is there a closed form solution in terms of

*n*or is an answer in terms of the prime factorisation acceptable?

just on the off chance, my answer in terms of the prime factorisaion is below :)

__Spoiler__:

apologies if that was really obvious to the rest of you and you wanted a function in terms of n. - Jul 5th 2011, 09:30 AMSpringFan25Re: Divisibility Probability
please can we see the solution now? :)

- Jul 5th 2011, 01:38 PMChris11Re: Divisibility Probability
Sorry for not replying earlier. My solution is actually the same as yours, I hope you enjoyed it. I thought it might have been a bit too easy.

- Jul 6th 2011, 11:19 AMSpringFan25Re: Divisibility ProbabilityQuote:

I thought it might have been a bit too easy.

PS: I have never done number theory though, i guess it might have been simpler if youve studied that properly ^^