It is a mabe a stupid question for many experts here. There is something wrong in the following reasoning, and now I could not find it. Could someone help me out? Any advice will be highly appreciated.
Denote $\tau(n)$ be the divisor function, and $p$ be a prime number. We can write L-series $$\sum_{n \ge 1 }\frac{\tau^2(n)} {n^s}=\sum_{(n,p)=1 }\frac{\tau^2(n)}{n^s}+\sum_{p |n }\frac{\tau^2(n)}{n^s},$$which is further transformed to $$\sum_{n \ge 1 }\frac{\tau^2(n)} {n^s}=\Big ( \sum_{n \ge 1 }\frac{\tau^2(n)} {n^s} \Big )\times \Big (\sum_{n |p^\infty}\frac{\tau^2(n)}{n^s}\Big )^{-1}+\sum_{p |n }\frac{\tau^2(n)}{n^s}. \tag{1}$$For the time being we get $$\Big \{ 1-\Big(\sum_{n |p^\infty}\frac{\tau^2(n)}{n^s}\Big)^{-1} \Big \}\times\sum_{n \ge 1 }\frac{\tau^2(n)} {n^s}=p^{-s}\sum_{n\ge 1 }\frac{\tau^2(np)}{n^s}. \tag{2}$$ Next we derive the exact formula for $\sum_{n\ge 1 }{\tau^2(np)}/{n^s}$. We have \begin{equation}\begin{split} \sum_{n\ge 1 }\frac{\tau^2(np)}{n^s}&=\tau^2(p) \sum_{(n,p)=1}\frac{\tau^2(n)}{n^s}+\sum_{p|n }\frac{\tau^2(np)}{n^s}\\ &=\tau^2(p) \Big(\sum_{n\ge1}\frac{\tau^2(n)}{n^s}-\sum_{p|n}\frac{\tau^2(n)}{n^s}\Big )+\sum_{p|n }\frac{\tau^2(np)}{n^s} . \end{split}\tag{3} \end{equation} By Hecke multiplicative relation，we have $\tau(pn)=\tau(n)\tau(p)-\tau(n/p)$ for any $p|n.$ Then we arrive at
\begin{equation}\begin{split} \sum_{n\ge 1 }\frac{\tau^2(np)}{n^s} =&\tau^2(p) \Big(\sum_{n\ge1}\frac{\tau^2(n)}{n^s}-\sum_{p|n}\frac{\tau^2(n)}{n^s}\Big )\\ &+\sum_{p|n }\frac{(\tau(n)\tau(p)-\tau(n/p))^2}{n^s} \\ =&\tau^2(p) \Big(\sum_{n\ge1}\frac{\tau^2(n)}{n^s}-p^{-s}\sum_{n \ge 1}\frac{\tau^2(np)}{n^s}\Big )\\ &+p^{-s}\sum_{n\ge 1 }\frac{(\tau(np)\tau(p)-\tau(n))^2}{n^s}.\end{split}\tag{4} \end{equation} By expending the square \begin{equation}\begin{split} \sum_{n\ge 1 }\frac{\tau^2(np)}{n^s} =&\tau^2(p) \sum_{n\ge1}\frac{\tau^2(n)}{n^s}+p^{-s}\sum_{n\ge 1} \frac{\tau^2(n)}{n^s}\\ &-2\tau(p)p^{-s}\sum_{n\ge 1} \frac{\tau(np)\tau(n)}{n^s}.\end{split}\tag{5} \end{equation} By (67) in the paper 'Mollification of the fourth moment of automorphic L-functions and arithmetic applications' or (61) in the paper 'Bounds for automorphic L-functions. II', we know that $$\sum_{n\ge 1} \frac{\tau(np)\tau(n)}{n^s}=2(1+p^{-s})^{-1}\sum_{n\ge 1} \frac{\tau^2(n)}{n^s}.$$ Hence we have by $(5)$ \begin{equation}\begin{split} \sum_{n\ge 1 }\frac{\tau^2(np)}{n^s} =&\Big \{4 +p^{-s}-8(1+p^{s})^{-1}\Big \}\sum_{n\ge1}\frac{\tau^2(n)}{n^s}.\end{split}\tag{6} \end{equation} We denote $p^{-s}=X $ for convenience. Combing $(6)$ and $(2)$, we get $$ 1-\Big(\sum_{n |p^\infty}\frac{\tau^2(n)}{n^s}\Big)^{-1}=p^{-s}\Big \{4 +p^{-s}-8(1+p^{s})^{-1}\Big \}=\frac{X(X^2-3X+4)}{1+X}.$$ It follows that $$\sum_{n |p^\infty}\frac{\tau^2(n)}{n^s}=\frac{1+X}{(1-X)^3}.$$
Writing $$\sum_{n |p^\infty}\frac{\tau^2(n)}{n^s}=\sum_{\alpha=0}^\infty \tau^2(p^{\alpha})X^{\alpha}=2\sum_{\alpha=0}^\infty (\alpha+1)X^{\alpha}=\frac{2}{(1-X)^2}.$$ Here we arrive at a contradiction！
I think that your mistake lies in the last line. We have $\tau^2(p^\alpha) = (\alpha+1)^2$, not $2(\alpha+1)$. And indeed, $$\sum_{\alpha\geq 0}(\alpha+1)^2X^\alpha = \frac{1}{X}\left(X \frac{d}{dX}\right)^2 \frac{X}{1-X} = \frac{X+1}{(1-X)^3}.$$