Implying form the Symmetry of the sum of von Mangoldt function

Question

In the article "On the Selberg-Erdos Proof of The Prime Number Theorem" by Ashvin A. Swaminathan, page 5, it is written -

By rearranging the claimed equality, it suffices to show that $$\sum_{\substack{mn≤x\\ m, n \; \text{not both prime}}} Λ(m)Λ(n) = O(x)$$ and BY SYMMETRY, it further suffices to show that $$\sum_{\substack{mn≤x\\ m\; \text{not prime}}} Λ(m)Λ(n) = O(x)$$

QUESTION:

1. How the symmetry is working here? What is the meaning of symmetry here?

2. How BY SYMMETRY, it is suffices to show that $\sum_{\substack{mn≤x\\ m\; \text{not prime}}} Λ(m)Λ(n) = O(x)$ from $\sum_{\substack{mn≤x\\ m, n \; \text{not both prime}}} Λ(m)Λ(n) = O(x)$?

Answer 1

The two-parametric function \begin{align*} \Phi_{m,n}(x)&=\sum_{{mn\leq x}\atop{m\,\text{not prime}}}\Lambda(m)\Lambda(n)\\ \end{align*} in $x$ is symmetric in the parameters $m$ and $n$ \begin{align*} \Phi_{m,n}(x)=\Phi_{n,m}(x) \end{align*}

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If we can show the validity of \begin{align*} \color{blue}{\sum_{{mn\leq x}\atop{m\,\text{not prime}}}\Lambda(m)\Lambda(n)=\mathcal{O}(x)}\tag{1} \end{align*} we have by symmetry (i.e. exchanging $m$ with $n$) \begin{align*} \sum_{{nm\leq x}\atop{n\,\text{not prime}}}\Lambda(n)\Lambda(m)=\mathcal{O}(x) \end{align*} from which \begin{align*} \sum_{{mn\leq x}\atop{m\,\text{not prime}}}\Lambda(m)\Lambda(n) +\sum_{{mn\leq x}\atop{n\,\text{not prime}}}\Lambda(m)\Lambda(n) =\mathcal{O}(x)+\mathcal{O}(x) =\mathcal{O}(x)\tag{2} \end{align*} follows.

Since \begin{align*} \sum_{{mn\leq x}\atop{m,n\,\text{not both prime}}}\Lambda(m)\Lambda(n) &\leq \sum_{{mn\leq x}\atop{m\,\text{not prime}}}\Lambda(m)\Lambda(n) +\sum_{{mn\leq x}\atop{n\,\text{not prime}}}\Lambda(m)\Lambda(n)\tag{$\ast$}\\ \end{align*} we have from (2) \begin{align*} \color{blue}{\sum_{{mn\leq x}\atop{m,n\,\text{not both prime}}}\Lambda(m)\Lambda(n)=\mathcal{O}(x)}\tag{3} \end{align*} Conclusion: The validity of (1) is sufficient by symmetry to show (3).

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Add-on: With respect to OP's question in the comment section we look at the inequality ($\ast$) and analyse when a strict inequality $(<)$ is given.

Given a sum in the form $\sum_{n\leq x}f(n)$ where $f$ is an arithmetical function, we consider $x$ to be a positive real number. So the sum is defined to be \begin{align*} \sum_{n\leq x}f(n):=\sum_{n=1}^{\lfloor x\rfloor}f(n) \end{align*} with $\lfloor x \rfloor$ being the integral part of $x$. Taking $x$ as a positive real number is convenient since then we can use for instance the big-O machinery for estimation. In case of $0<x<1$ the sum is the empty sum and set to $0$. See e.g. sections 3.1 and 3.2 in Introduction to Analytic Number Theory by T.M. Apostol.

The Von Mangoldt function $\Lambda(n)$ takes non-negative values only, so that \begin{align*} \sum_{mn\leq x}\Lambda(m)\Lambda(n)\tag{$\ast\ast$} \end{align*} is monotonically increasing with $x$.

When going through small pairs $(m,n)$ with increasing sum $m+n\geq 1$ we have to find out the first occurrence where both values $m,n$ have a form $p^k$ with $p$ prime and $k>1$. Because this is the first time where both sums of the RHS of ($\ast$) provide a non-negative contribution which is then twice the contribution of the LHS.

The first pair is $(m,n)=(4,4)$ and we conclude due to monotonicity of ($\ast\ast$) that equality is given for $\color{blue}{0<x<16}$ and strict inequality for $\color{blue}{x\geq 16}$.