finding the number of positive divisors for a $1111.......1$ ($1992$ times)

Question

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 $\underbrace{111\ldots1}_{1992 \text{ times}}$. 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?

Answer 1

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.