<s>[INST] Computing $\prod_p(\frac{p^2-1}{p^2+1})$ without the zeta function? We see that $$\frac{2}{5}=\frac{36}{90}=\frac{6^2}{90}=\frac{\zeta(4)}{\zeta(2)^2}=\prod_p\frac{(1-\frac{1}{p^2})^2}{(1-\frac{1}{p^4})}=\prod_p \left(\frac{(p^2-1)^2}{(p^2+1)(p^2-1)}\right)=\prod_p\left(\frac{p^2-1}{p^2+1}\right)$$
$$\implies \prod_p \left(\frac{p^2-1}{p^2+1}\right)=\frac{2}{5},$$
But is this the only way to compute this infinite product over primes? It seems like such a simple product, one that could be calculated without the zeta function.
Note that $\prod_p(\frac{p^2-1}{p^2+1})$ also admits the factorization $\prod_p(\frac{p-1}{p-i})\prod_p(\frac{p+1}{p+i})$. Also notice that numerically it is quite obvious that the product is convergent to $\frac{2}{5}$: $\prod_p(\frac{p^2-1}{p^2+1})=\frac{3}{5} \cdot \frac{8}{10} \cdot \frac{24}{26} \cdot \frac{48}{50} \cdots$. 
 [/INST] I found a proof of 
$$5 \sum_{m=1}^{\infty} \frac{1}{m^4} = 2 \left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right)^2$$
by rearranging sums and wrote it up. The argument is just 1.5 pages, the other 4.5 are explanations and context.
Here is a summary using divergent sums; see the write up for a correct version. Set
$$h(m,n) = \begin{cases}
\frac{1}{m^3 (n-m)} & m \neq n,\ m \neq 0, \\
0 & m=n \ \text{or} \ m=0.
\end{cases}$$
Then we should have
$$\sum_{(m,n) \in \mathbb{Z}^2} h(m,n) - h(n,2n-m) =0$$
as every value occurs twice with opposite signs. So $\sum_{(m,n) \in \mathbb{Z}^2} g(m,n)=0$ where
$$g(m,n) :=  h(m,n) - h(n,2n-m) =   \begin{cases}
\frac{m^2+mn+n^2}{m^3 n^3} & m \neq n,\ m, n \neq 0 \\
- \frac{1}{m^4} & n=0,\ m \neq 0 \\
- \frac{1}{n^4} & m=0,\ n \neq 0 \\
0 & m=n \end{cases}$$
Group together the terms where $(|m|, |n|)$ have a common value; we get $\sum_{(m,n) \in \mathbb{Z}_{\geq 0 }^2} f(m,n) =0$ where
$$f(m,n) = \begin{cases} 
\frac{4}{m^2 n^2} & m \neq n,\ m,n >0, \\
- \frac{2}{m^4} & m>n=0, \\
- \frac{2}{n^4} & n>m=0, \\
- \frac{2}{m^4} & m=n>0, \\
0 & m=n=0.
 \end{cases} $$
Writing this out, $4\zeta(2)^2 - 6 \zeta(4) - 4 \zeta(4)=0$, as desired.
Has anyone seen this? If this is new, I'm thinking of sending it to the Monthly.
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