Divisor squared?

Consider $N = p_{1}^{k_1}p_{2}^{k_2}...p_{j}^{k_j}$ as in canonical representation, then it has $\tau(N) = (k_{1} +1)(k_{2} +1)...(k_{j} +1)$ divisors. We have $N^2 = (p_{1}^{k_1}p_{2}^{k_2}...p_{j}^{k_j})^2 = p_{1}^{2k_1}p_{2}^{2k_2}...p_{j}^{2k_j}$ which has $\tau(N^2) = (2k_{1} +1)(2k_{2} +1)...(2k_{j} +1)$ divisors. Now, this means that if we were to check for the possible values for $\tau(N^2)$ given $\tau(N)$ , then we just need to add each of the $k_i$ 's to each of the respective product term's $(k_i + 1)$ in $\tau(N)$ .

To do this, we are going to try to split 18 into products of terms, then using the "split" 18s, we can compare terms-by-terms correspondence with $\tau(N)$ , then use that in order to form all possible values of $\tau(N^2)$ , given that $\tau(N) = 18$ .

There are a total of 4 ways to factorize 18:

$18 = 2 \times 3 \times 3 = 2 \times 9 = 6 \times 3 = 1 \times 18$

We are going to check the solutions for $\tau(N) = 18$ :

Case 1 : $\tau(N) = 2 \times 3 \times 3$

We have that $(k_{1} +1)(k_{2} +1)(k_{3} +1) = 2 \times 3 \times 3 \Rightarrow (2k_1 + 1)(2k_2 +1)(2k_3 + 1) = 3 \times 5 \times 5 = 75 = \tau(N^2)$

Hence $\tau(N^2)$ can be $75$ .

Case 2 : $\tau(N) = 2 \times 9$

We have that $(k_{1} +1)(k_{2} +1) = 2 \times 9 \Rightarrow (2k_1 + 1)(2k_2 +1) = 3 \times 17 = 51 = \tau(N^2)$

Hence $\tau(N^2)$ can be $51$ .

Case 3 : $\tau(N) = 6 \times 3$

We have that $(k_{1} +1)(k_{2} +1) = 6 \times 3 \Rightarrow (2k_1 + 1)(2k_2 +1) = 11 \times 5 = 55 = \tau(N^2)$

Hence $\tau(N^2)$ can be $55$ .

Case 4 : $\tau(N) = 1 \times 18$

We have that $k_{1} +1 = 18 \Rightarrow 2k_1 + 1 = 35 = \tau(N^2)$

Hence $\tau(N^2)$ can be $35$ .

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In conclusion, the possible values of $\tau(N^2)$ are ${35, 55,51,75 }$ . We see that $95$ is not in the list.

Therefore $\boxed{95}$ can not be the total number of positive divisors of $N^2$ .