Can you prove or disprove the following claim:


First, define the function $\xi(n)$ as follows: $$\xi(n)=\begin{cases}-1, & \text{if }\varphi(n) \equiv 0 \pmod{4} \\
1, & \text{if }\varphi(n) \equiv 2 \pmod{4} \\ 0, & \text{if otherwise }
\end{cases}$$
where $\varphi(n)$ denotes [Euler's totient function](https://en.wikipedia.org/wiki/Euler%27s_totient_function). Then,
$$\frac{\pi^2}{72}=\displaystyle\sum\_{n=1}^{\infty}\frac{\xi(n)}{n^2}$$


The SageMath cell that demonstrates this claim can be found [here](https://sagecell.sagemath.org/?z=eJwrTbI1NQACa66KTI08TdvqzDSN1NKc1KKCDBBf1cTW1kgn2dZQB1PcACiuawgkDDQ1rYtSS0qL8jSSNa1ruYpti0tzNfKAmkqTdMDG6mvkxRkBVXEVFGXmlWgUaxnqwTkBmXFG-uZGmtYAT1go5g==&lang=gp&interacts=eJyLjgUAARUAuQ==).



An infinite series involving the mod-parity of Euler's totient function
The only odd values of $\phi(n)$ are $\phi(1)=\phi(2)=1$.


$\phi(n)$ is even but not divisible by $4$ when:


1. $n=4$
2. $n=2^{\left\{0,1\right\}}p^m$, where $p=4k+3$ is prime, $m=1,2,3,...$


We have
$$
\frac{\pi^2}{6}=1+\frac14+\sum\_{\substack{n=1\\\phi(n)\equiv 0}}^\infty\frac{1}{n^2}+\sum\_{\substack{n=1\\\phi(n)\equiv 2}}^\infty\frac{1}{n^2}.
$$
(congruences are modulo $4$)


The claim in the question reads
$$
\frac{\pi^2}{72}=-\sum\_{\substack{n=1\\\phi(n)\equiv 0}}^\infty\frac{1}{n^2}+\sum\_{\substack{n=1\\\phi(n)\equiv 2}}^\infty\frac{1}{n^2}.
$$


Combining these two we get a hypothetical identity
$$
\sum\_{\substack{n=1\\\phi(n)\equiv 2}}^\infty\frac{1}{n^2}=\frac{13\pi^2}{144}-\frac{5}{8}.
$$
However
$$
\sum\_{\substack{n=1\\\phi(n)\equiv 2}}^\infty\frac{1}{n^2}=\frac{1}{16}+\left(1+\frac14\right)\sum\_{m=1}^\infty\sum\_{p\equiv 3}\frac{1}{p^{2m}}=\frac{1}{16}+\frac54\sum\_{p\equiv 3}\frac{1}{p^2-1}.
$$
Thus the conjecture is equivalent to
$$
\sum\_{p\equiv 3}\frac{1}{p^2-1}=\frac{1}{20} \left(\frac{13 \pi ^2}{9}-11\right).\tag{\*}
$$
The article on sums over primes on mathworld <https://mathworld.wolfram.com/PrimeSums.html> does not list any sums of this kind. Numerical checks show that the claim is false. For example summing over the first $N=500000$ primes gives for the ratio of the LHS of (\*) to the RHS:
$$
1.000153116
$$
with the truncation error term of the order
$$
\int\_{N}^\infty \frac{dx}{x^2\ln^2(x)}\sim\frac{1}{N\ln^2(N)}\sim 10^{-10}.
$$


