$P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-...}}}}$ Vassilev-Missana - A note on prime zeta function and Riemann zeta function¹ claims the following remarkable identity:
$$
P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-...}}}},
$$
for integer $s>1$, where $\zeta(s)$ is the Riemann zeta function defined as $\sum_{n=1}^\infty \frac{1}{n^s}$ and $P(s)$ denotes the prime zeta function $\sum_{p\in\mathbb{P}}^\infty \frac{1}{p^s}$, and $\mathbb{P}$ is the set of prime numbers.
I am not a number theoretician, but the paper looks like it might contain some errors. I have already raised the question here, and it looks like I may be right. However, after performing some calculations using Wolfram Alpha, it looks like the identity might have some truth to it.
If we denote by $\epsilon(s)$ the error term $P(s)-\left(1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-...}}\right)$, we get the following results for successive finite approximations to the nested radical, in which the $k^{\rm th}$ approximation is $\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{...-\sqrt{\frac{2}{\zeta(2^{k-1}s)}}}}}$, and in this case by taking $s=2$:
\begin{array}{|c|c|}
\hline
k & \epsilon(2)\\ \hline
3 &  0.19732\ldots \\ \hline
4 &  -0.13198\ldots \\ \hline
6 &  0.04839\ldots \\ \hline
5 &  -0.035665\ldots \\ \hline
7 &  0.007512\ldots \\ \hline
8 & -0.013753\ldots \\ \hline
9 & -0.00304\ldots \\ \hline
\end{array}
so indeed it looks like $\epsilon(2)$ is converging to $0$ (note the expression "looks like", as I have not run any calculations for really large $k$).
Additionally, by taking $s=3$ the identity above would remarkably imply that:
$$
\zeta(3) = 2\left((1-P(3))^2+\sqrt{\frac{2}{\zeta(6)}-\sqrt{\frac{2}{\zeta(12)}-\sqrt{\frac{2}{\zeta(24)}-...}}}\right)^{-1},
$$
and indeed taking the finite nested radical approximation for $k=9$, the error term is already $\epsilon=-0.002516\ldots$ and seems to approach $0$ as $k\to\infty$ (note again the expression "seems to").
Is it possible that the identity still holds, nonwithstanding the issues in the paper, and if so, how may one prove it? Or do the approximations for  large $k$ actually not produce $\epsilon\to0$? If so, how can the apparent convergence be explained?

¹ Note that the journal in question is a reputed one, with Paul Erdős being at some point the editor (between '95 and '96), not a second-rate journal.
 A: This has been answered in the comments by Lucia. The identity $$P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-\cdots}}}}$$ is false. By subtracting $1$ from both sides and squaring, we have that $$(1-P(s))^{2}+1-\frac{2}{\zeta(s)}=1-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-\cdots}}},$$ which by the identity implies that $$(1-P(s))^{2}+1-\frac{2}{\zeta(s)}=P(2s).$$ This equation, which appears as theorem $1$ in the paper, is not true.
A: This so-called "functional equation"
$$(1-P(s))^{2}+1-\frac{2}{\zeta(s)}=P(2s)$$
which was published in a 2016 paper has been used in a 2019 paper claiming to prove the generalized Riemann hypothesis. As all here have pointed out, that equation is false. A disproof has been recently published in 2021 and can be found here.  It interested me to discover this, because in 2015 I had derived this same false equation. You can find my posts about it here and here. I wanted to take it upon myself to show a generalization that highlights the error.
There is a formula that relates the partial euler prouct to the partial prime zeta function.
$$\prod_{k=1}^n\left(1-\frac{1}{{p_k}^s}\right)=\sum_{r=1}^n\sum_{\{r\}}\prod_{j=1}^l\frac{(-1)^{i_j}}{i_j!}\left(\frac{{P_n(s\cdot r_j)}}{r_j}\right)^{i_j}$$
where the sum is over partitions of $r$ and $P_n(s)=\sum_{k=1}^n\frac{1}{{p_k}^s}$, the partial prime zeta function. $p_*\in\text{Primes}$ (Note: this formula holds if we use any set of real numbers, not just the primes). For more details see Grunberg's paper On asymptotics, Stirling numbers, Gamma function, and polylogs
An equivalent representation is
$$
\prod_{k=1}^n\left(1-\frac{1}{{p_k}^s}\right)=\sum_{k=0}^n{\frac{B_k(-0!P_n(s),-1!P_n(2s),\dots,-(k-1)!P_n(ks))}{k!}}
$$
where $B_n(x_1,\dots,x_n)$ is the complete exponential Bell polynomial.
On a separate note, this also is equal to the analytic form, $\exp{\left(-\sum_{k=1}^\infty{\frac{P_n(ks)}
{k}}\right)}$
This means we can write
$$
\begin{align}
\left(1-\frac{1}{2^{s}}\right)
&=1-P_1\left(s\right)
\\
&=\color{red}{1-P_1\left(s\right)}\underbrace{\color{red}{-\frac{P_1\left(2s\right)}{2}+\frac{P_1\left(s\right)^{2}}{2}}}_0
\\
&=1-P_1\left(s\right)\underbrace{-\frac{P_1\left(2s\right)}{2}+\frac{P_1\left(s\right)^{2}}{2}}_0\space\underbrace{-\space\frac{P_1\left(3s\right)}{3}+\frac{P_1\left(2s\right)P_1\left(s\right)}{2}-\frac{P_1\left(s\right)^{3}}{6}}_0
\\
&=\cdots
\\
\left(1-\frac{1}{2^{s}}\right)\left(1-\frac{1}{3^{s}}\right)
&=\color{red}{1-P_2\left(s\right)-\frac{P_2\left(2s\right)}{2}+\frac{P_2\left(s\right)^{2}}{2}}
\\
&=1-P_2\left(s\right)-\frac{P_2\left(2s\right)}{2}+\frac{P_2\left(s\right)^{2}}{2}\underbrace{-\frac{P_2\left(3s\right)}{3}+\frac{P_2\left(2s\right)P_2\left(s\right)}{2}-\frac{P_2\left(s\right)^{3}}{6}}_0
\\
&=\cdots
\\
\left(1-\frac{1}{2^{s}}\right)\left(1-\frac{1}{3^{s}}\right)\left(1-\frac{1}{5^{s}}\right)
&=1-P_3\left(s\right)-\frac{P_3\left(2s\right)}{2}+\frac{P_3\left(s\right)^{2}}{2}-\frac{P_3\left(3s\right)}{3}+\frac{P_3\left(2s\right)P_3\left(s\right)}{2}-\frac{P_3\left(s\right)^{3}}{6}
\\
&=\cdots
\end{align}
$$
and so on.
In my case, I derived those lines with the red RHS in a different way and falsely assumed that the number of terms in the product on the LHS, as well as the number of summands in the partial prime zeta functions on the RHS could be continued beyond two terms. This is where the Riemann zeta function and the so-called "functional equation" came from.
These equations are true for any set of real numbers not just powers of primes starting from $2$. That's why as Lucia notes that the 2016 “identity” is true only if the Riemann zeta (as an Euler product) and the prime zeta functions involve a single prime term or two prime terms.
Thus a true generalized identity analogous to the false one is
$$\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)=1-\left(\frac{1}{a}+\frac{1}{b}\right)-\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)}{2}+\frac{\left(\frac{1}{a}+\frac{1}{b}\right)^{2}}{2}$$
from which we can derive the nested radical identity
$$1-\frac{1}{a}-\frac{1}{b}=\\
\lim\limits_{n\to\infty}\sqrt{2\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)-\sqrt{2\left(1-\frac{1}{a^{2}}\right)\left(1-\frac{1}{b^{2}}\right)-\cdots-\sqrt{2\left(1-\frac{1}{a^{2^n}}\right)\left(1-\frac{1}{b^{2^n}}\right)-1}}}$$
