Welcome 2016!

If a natural number $n$ has prime factorization $p_1^{e_1} p_2^{e_2} p_3^{e_3} \ldots$ , then the number of positive divisors it has is $(1+e_1)(1+e_2)(1+e_3)\ldots$ . This can be proven as follows: clearly all divisors $d$ of $n$ are in the form $p_1^{f_1} p_2^{f_2} p_3^{f_3} \ldots$ (there can be no prime that divides $d$ but not $n$ ), and we also have $0 \le f_i \le e_i$ for all $i$ (if $f_i > e_i$ , then $p_i^{f_i}$ divides $d$ but not $n$ ). Other than that, any such choice for $f_1, f_2, f_3, \ldots$ works. There are $1+e_i$ choices for $f_i$ , so we multiply everything to get $(1+e_1)(1+e_2)(1+e_3)\ldots$ .

Applying this to $n = 2016^2 = 2^{10} \cdot 3^4 \cdot 7^2$ gives $(1+10)(1+4)(1+2) = 11 \cdot 5 \cdot 3 = 165$ positive divisors.

To figure out the amount of factors one number has, add 1 to each exponent when prime factorized and multiply the exponents together. (Here, it's 2^10 x 3^4 x 5^2.. so add 1 to each exponent so 11, 5, 3 and multiply together)

For better understanding, I edited the solution a bit.Thanks.

One of $x,y$ can be negative (and the other will be positive). For example, $x = 1008, y = -2016$ is a solution.