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.
 A: Reinforcing Nemo's negative answer.
$$  \sum_{n=1}^{10^4} \frac{\xi(n)}{n^2} - \sum_{n=10^4+1}^\infty \frac{1}{n^2} \leq \sum_{n=1}^{\infty} \frac{\xi(n)}{n^2} \leq \sum_{n=1}^{10^4} \frac{\xi(n)}{n^2} + \sum_{n=10^4+1}^\infty \frac{1}{n^2}  $$
The first and third of these are directly computable:
$$  0.13712{\dots} \leq \sum_{n=1}^{\infty} \frac{\xi(n)}{n^2} \leq 0.13732\dots  \text{.}  $$
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.)
A: The only odd values of $\phi(n)$ are $\phi(1)=\phi(2)=1$.
$\phi(n)$  is even but not divisible by $4$ when:

*

*$n=4$


*$n=2^{\left\{0,1\right\}}p^m$, 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
$$
\int_{N}^\infty \frac{dx}{x^2\ln^2(x)}\sim\frac{1}{N\ln^2(N)}\sim 10^{-10}.
$$
