, 2016

A number with prime factorization $p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k}$ has
$$\tau(p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k}) = (\alpha_1 + 1)(\alpha_2 + 1) \dots (\alpha_k + 1)$$
divisors. The number $N$ can be factored as $N = 2^{15} \cdot 5 \cdot 13 \cdot 31$, so
$$\tau(N) = (15 + 1) \cdot (1 + 1) \cdot (1 + 1) \cdot (1 + 1) = 2^7 = 128$$
$$\text{and } \tau(N^2) = (2 \cdot 15 + 1) \cdot (2 \cdot 1 + 1) \cdot (2 \cdot 1 + 1) \cdot (2 \cdot 1 + 1) = 31 \cdot 3^3 = 837.$$
If $d$ divides $N^2$ then $\frac{N^2}{d}$ also divides $N^2$, so all divisors can be sorted into pairs $(d, \frac{N^2}{d})$, omitting the number $N$, which would otherwise be paired with itself. In each pair exactly one divisor is smaller than $N$ and one is greater than $N$. This implies that the number of divisors of $N^2$ which are strictly smaller than $N$ is equal to the number of pairs, which is
$$\frac{\tau(N^2) - 1}{2}.$$

Each divisor of $N$ also divides $N^2$, so we have to subtract the number of divisors of $N$ smaller than $N$. The number of divisors of $N^2$ which are smaller than $N$ and do not divide $N$ is therefore equal to
$$\frac{\tau(N^2) - 1}{2} - (\tau(N) - 1) = \frac{836}{2} - 127 = 291.$$