An infinite series involving the mod-parity of Euler's totient function

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. 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.

• Strange choice of tags... Commented Aug 29, 2021 at 8:39

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 $$\int_{N}^\infty \frac{dx}{x^2\ln^2(x)}\sim\frac{1}{N\ln^2(N)}\sim 10^{-10}.$$

$$\sum_{n=1}^{10^4} \frac{\xi(n)}{n^2} - \sum_{n=10^4+1}^\infty \frac{1}{n^2} \leq \sum_{n=1}^{\infty} \frac{\xi(n)}{n^2} \leq \sum_{n=1}^{10^4} \frac{\xi(n)}{n^2} + \sum_{n=10^4+1}^\infty \frac{1}{n^2}$$
$$0.13712{\dots} \leq \sum_{n=1}^{\infty} \frac{\xi(n)}{n^2} \leq 0.13732\dots \text{.}$$ However, $$\frac{\pi^2}{72} = 0.137077{\dots} \text{,}$$ which does not fit between those bounds.
(Aside: There may be smaller cutoffs than $$10^4$$ that exhibit this negative result. $$10^3$$ does not.)