So the actual question is to prove that the number of positive divisors is even. But to do that I have to find the number of positive divisors for 111.....1(1992 1's). I know that I should try to find the prime factorisation and then use combinatorics from there , but I don't even know how to factorize such a large number. Is my approach to this question wrong?

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    $\begingroup$ $111$...$1$ ($1992$ $1$'s) sums to $1992$ which is divisible by $3$ but not by $9$. Therefore $111$...$1$ is not a square number. $\endgroup$ – Mohammad Zuhair Khan Aug 20 '18 at 3:06

No, you don't have to find the number of divisors nor the prime factorization. Hint:

For each divisor $d$ of $n$, there is another divisor $\frac{n}{d}$. So, the divisors come in pairs. When is the number of divisors odd? Well, whenever $d=\frac nd$ occurs, meaning, $n$ is a square.

Two examples: Take $n=12$; take a divisor, say, $2$, and see that $\frac{12}2=6$ is also a divisor. So we get the pairs $(1,12), (2,6), (3,4)$.

The second example, when this doesn't work, is $n=16$; we get the pairs $(1,16),(2,8)$, but $4$ is on its own since $\frac{16}4=4$.

So, all you have to do is show $111\cdots 1$ is not a square.

And as a sidenote, I'm fairly sure you're not supposed to be trying to find its prime factorization. Especially not by hand.

  • $\begingroup$ hmm why what do you mean the divisors come in pairs? for n = 12 do u mean (6,2) and (12,1) ? $\endgroup$ – calveeen Aug 20 '18 at 2:51
  • $\begingroup$ Or how did you get (3,4) as one of the pairs as well? from d = n/d $\endgroup$ – calveeen Aug 20 '18 at 2:52
  • $\begingroup$ Well, $3$ is a divisor, so we have $12/3 = 4$ as well $\endgroup$ – vrugtehagel Aug 22 '18 at 1:02

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