Number of positive divisor Given the following number "$11..11$" -$1992$ ones repeated-, prove that the number of positive divisor is even.
I came up with the following idea: pick the number and rewrite the number this way:
$$\sum_{i=0}^{1992}{10^i}=\sum_{i=0}^{1992}{2^i5^i}=2^05^0+2^15^1+\cdots+2^{1992}5^{1992}$$
And, the only common divisor to all those number is $1$ and the big number itself, only two divisors. So the amount of positive divisors of the whole number must be even.
Does this make any sense?
Thanks you
 A: Fact 1: Every integer squared is congruent to either $0$ or $1$ modulo $4$.
Fact 2: $1\ldots1 \equiv 3 \text{ mod } 4$. 
Therefore, $1 \ldots 1$ is not a square. QED

In case the two facts here are unclear or unknown, let us establish each:
Fact 1: Any number is $0, 1, 2,$ or $3$ modulo $4$; squaring and reducing modulo $4$ again, the results are $0, 1, 0, 1$, respectively. Thus, the first fact is established.
Fact 2: Observe $11 = 8 + 3$, and $111 = 108 + 3$, and $1111 = 1108 + 3$, etc. Looking at the right hand side in each case, the former addend is divisible by $4$, and the $3$ stands alone. Thus, reducing modulo $4$ establishes the second fact.
A: Hint For any integer $n$ if $d$ is a divisor of $n$ then so is $\frac{n}{d}$. As long as $\frac{n}{d}\neq d$ they are a pair, and you can pair the divisors.
Use this idea to show that any number which is NOT a perfect square has an even number of divisors.
Hint 2
$$111...1=\frac{1}{9} 999..999=\frac{1}{9}(10^{1992}-1)$$
Use the fact that $10^{1992}$ is a perfect square to show that your number cannot be a perfect square.
A: The number $11..11$ is divisible by $3$ since the sum of its digits is $1+1+...=1992$=multiple of $3$ but not by $9$ since the sum of its digits is not divisible by $9$.
So, the number you investigate is not a perfect square  and therefore it has an even number of divisors.
A: Here's another modular proof. Divide the number by $11$ to get $$\underbrace{1010\ldots101}_{\textrm{996 1s}}.$$ We will show that this result is not itself divisible by $11$, implying that the original number is not a perfect square.
Note that a number with digits $a_1b_1a_2b_2 \ldots a_{k-1}b_{k-1}a_k$ is congruent to $0$ modulo $11$ (i.e., is divisible by $11$) if and only if $\sum a_i \equiv \sum b_i \pmod {11}$. In our case, we have $\sum a_i = 996$ and $\sum b_i = 0$. But $996 \not \equiv 0 \pmod {11}$.
