Proof conclusion to: Let m,n $\in$ $\mathbb{N}$ $\backslash$ $\{0\}$ . Prove that $s |\tau (m^{s}) - \tau(n^{s})$, where $\tau$ is a divisor function.
Hello, so I have a tiny question about the summarization of my proof. I got to the point where I defined my $\tau$ divisor function and got them both $\tau(m^s)$= $\prod$ something +1 and $\tau(n^{s})$ = $\prod$ something +1.
I've noticed that the last terms of the polynomial are both one. Why is it with the +1 part, I can conclude with that: s| $\tau(m^s)$ - $\tau(n^s)$.
The line for why that is, isn't quite clear.
Let $m=p_1^{a_1}p_2^{a_2}\dotsb p_k^{a_k}=\prod_{i=1}^k p_i^{a_i}$ and $n=\prod_{j=1}^t q_j^{b_j}$ be prime factorizations of $m$ and $n$. Then by definition, $$\tau(m^s)=\prod_{i=1}^k(a_is+1)=P(s)+1,$$ where $P(s)$ is a polynomial in $s$ with constant term $0$. Likewise $$\tau(n^s)=\prod_{j=1}^t(b_js+1)=Q(s)+1,$$ where $Q(s)$ is a polynomial in $s$ with constant term $0$.
Now $$\tau(m^s)-\tau(n^s)=P(s)-Q(s)=s(\text{some polynomial}).$$ Thus $s$ divides the difference.