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Let $\tau(n)$ denotes the number of divisors of $n$. Let $n,m,s,t\in \mathbb{N}$. Then prove that

  1. $(s-t)| \tau(n^s)-\tau(n^t)$
  2. $s|(\tau(m^s)-\tau(n^s)$

I tried it :

Let $$n=\prod_{i=1}^k p_i^{e_i}\implies n^s=\prod_{i=1}^kp_i^{e_i s}\ \text{and } n^t=\prod_{i=1}^kp_i^{e_i t}.$$ Now how will I show that $(s-t)|\tau\left(\prod_{i=1}^kp_i^{e_i s}\right)-\tau\left(\prod_{i=1}^kp_i^{e_i t}\right)$

As $\tau$ is a multiplicative function so we can write $\tau\left(\prod_{i=1}^kp_i^{e_i s}\right)-\tau\left(\prod_{i=1}^kp_i^{e_i t}\right)=\prod_{i=1}^k\tau(p_i^{e_i s})-\prod_{i=1}^k\tau(p_i^{e_i t})=\prod_{i=1}^k(1+e_is)-\prod_{i=1}^k(1+e_it)$. Then I stuck.

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  • $\begingroup$ primes.utm.edu/glossary/xpage/tau.html $\endgroup$ Oct 25, 2017 at 14:48
  • $\begingroup$ Yeah, basically I forgot to type these steps at that moment I was thinking that I have typed but when I saw it again I found I have left it halfway only. $\endgroup$
    – I am pi
    Oct 25, 2017 at 14:52

2 Answers 2

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Hint: 1. Let $f(s)=\tau(n^s)-\tau(n^t)$, $t$ fixed. It is easy to see that $f(s)$ is a polynomial of the variable $s.$ Since $f(t)=0$ by Bezout's theorem it implies that $s-t$ divides $f(s)$.

  1. The last terms of the polynomials $\tau(n^s),\tau(m^s)$ are equal to $1.$ Thus their difference has not the last term and then $s$ divides the difference.
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Hint: $$\prod_{i=1}^k\tau(p_i^{e_i s})-\prod_{i=1}^k\tau(p_i^{e_i t}) = \prod_{i=1}^k \tau(p_i^{e_i t})\left(\prod_{i=1}^k\tau(p_i^{e_i (s-t)})-1\right)$$

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