Riemann Zeta Function of 4/Proof 5

Proof
Create a multiplication table where the column down the and the row across the top each contains the terms of Zeta of two


 * $\begin{array}{r|cccccccccc}

\displaystyle \paren {\map \zeta 2}^2 & \paren {\dfrac {1} {1^2}} & \paren {\dfrac {1} {2^2}} & \paren {\dfrac {1} {3^2}} & \paren {\dfrac {1} {4^2}} & \cdots \\ \hline

\paren {\dfrac {1} {1^2}} & \paren {\dfrac {1} {1^4}} & \paren {\dfrac {1} {1^2}} \paren {\dfrac {1} {2^2}} & \paren {\dfrac {1} {1^2}} \paren {\dfrac {1} {3^2}} & \paren {\dfrac {1} {1^2}} \paren {\dfrac {1} {4^2}} & \cdots \\

\paren {\dfrac {1} {2^2}} & \paren {\dfrac {1} {2^2}} \paren {\dfrac {1} {1^2}} & \paren {\dfrac {1} {2^4}} & \paren {\dfrac {1} {2^2}} \paren {\dfrac {1} {3^2}} & \paren {\dfrac {1} {2^2}} \paren {\dfrac {1} {4^2}} & \cdots \\

\paren {\dfrac {1} {3^2}} & \paren {\dfrac {1} {3^2}} \paren {\dfrac {1} {1^2}} & \paren {\dfrac {1} {3^2}} \paren {\dfrac {1} {2^2}} & \paren {\dfrac {1} {3^4}} & \paren {\dfrac {1} {3^2}} \paren {\dfrac {1} {4^2}} & \cdots \\

\paren {\dfrac {1} {4^2}} & \paren {\dfrac {1} {4^2}} \paren {\dfrac {1} {1^2}} & \paren {\dfrac {1} {4^2}} \paren {\dfrac {1} {2^2}} & \paren {\dfrac {1} {4^2}} \paren {\dfrac {1} {3^2}} & \paren {\dfrac {1} {4^4}} & \cdots \\

\cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ \end{array}$

The sum of all of the entries in this table is equal to $\paren {\map \zeta 2}^2$


 * $\map \zeta 4$ is the sum of the entries along the main diagonal

We have:

Therefore: