Relationship between the Number of Divisor

Question

If i know the divisor of number N let say it has X no. of divisor, then can i directly find the number of divisor of N^2,N^3,N^4...so on

Is there any relationship between them ?

Answer 1

Let $N = p_1^{k_1} p_2^{k_2} \ldots p_n^{k_n}$, where $p_i$ are distinct primes.

The number of factors of $N$ is $(k_1+1)(k_2+1)\ldots(k_n+1)$.

Now, suppose you take $N^s$. Then $N^s = p_1^{sk_1} p_2^{sk_2} \ldots p_n^{sk_n}$.

Hence the number of factors of $N^s$ is $(sk_1+1)(sk_2+1)\ldots(sk_n+1)$.

For example, suppose we have $12 = 2^1 3^2$. The number of factors of $12$ is $(1+1)(2+1)=6$, and the factors are $1,2,3,4,6,12$.

Now, for $12^2=144$, we say the number of factors are: $((1*2) + 1)((2*2) +1) = 15$. Indeed, $144$ has the factors $1,2,3,4,6,8,9,12,16,18,24,36,48,72,144$, which are $15$ of them. Similarly, $1728=12^3$ has $((3*1) + 1) ((3*2)+1) = 28$ factors (you can check them).

Answer 2

If I have understood your question, you are asking "does the value of $\sigma_0 (n)$ determine $\sigma_0(n^2)$?" . The answer to that question is No.

Here, as is standard, we are denoting the number of divisors of a natural number $n$ as $\sigma_0(n)$.

Recall that $n=\prod p_i^{a_i}\implies \sigma_0(n)=\prod (a_i+1)$.

We now exploit the fact that $12=3\times 4=6\times 2$.

Choose two distinct primes $p,q$ and let $n=p^3q^2,m=p^5q$. Then $\sigma_0(n)=4\times 3=12$ and $\sigma_0(m)=6\times 2=12$. However $$\sigma_0(n^2)=\sigma_0(p^6q^4)=7\times 5 =35$$ and $$\sigma_0(m^2)=\sigma_0(p^{10}q^2)=11\times 3 =33$$ So we have an infinite family of counterexamples.

Answer 3

If $x$ is odd or if $x=2$, then $N = p^{x-1}$ for some prime $p$ and then the number of divisors of $N^t$ is given directly $tx-t+1$. Otherwise there are several possibilities and you need to factor $N$ to know the correct one.