[proofplan]
Sieve the one-parameter Goldbach sequence given by the two linear forms $n$ and $N-n$. The local densities have dimension two away from prime divisors of $N$, and the standard Selberg upper-bound sieve gives a sifted-set bound with the Goldbach singular series. The prime representations lie in that sifted set apart from an $O(N^{1/2})$ endpoint contribution, which is absorbed because the singular series is bounded below on even $N$.
[/proofplan]
[step:Compute the local density for the Goldbach sieve]
For $2\le n\le N-2$, consider the pair of linear forms $n$ and $N-n$. For an odd prime $p$, the congruence
\begin{align*}
p\mid n(N-n)
\end{align*}
excludes the residue classes $n\equiv0\pmod p$ and $n\equiv N\pmod p$. If $p\nmid N$, these two classes are distinct, so $\omega_N(p)=2$. If $p\mid N$, they coincide, so $\omega_N(p)=1$. The prime $2$ is harmless because $N$ is even and contributes only to the fixed leading factor in $\mathfrak S(N)$.
Thus the squarefree local density is multiplicative:
\begin{align*}
\omega_N(d):=\prod_{p\mid d}\omega_N(p)
\end{align*}
for squarefree odd $d$. The elementary residue count gives
\begin{align*}
\#\{2\le n\le N-2:d\mid n(N-n)\}=\frac{\omega_N(d)}{d}N+O(\omega_N(d)).
\end{align*}
[guided]
The two forbidden conditions modulo an odd prime $p$ are $n\equiv0\pmod p$ and $N-n\equiv0\pmod p$, which is the same as $n\equiv N\pmod p$. If $p$ does not divide $N$, these are two different residue classes, so $\omega_N(p)=2$. If $p$ divides $N$, both forbidden conditions are the same class, so $\omega_N(p)=1$.
For squarefree odd $d$, the [Chinese remainder theorem](/theorems/734) makes these local choices independent, giving
\begin{align*}
\omega_N(d):=\prod_{p\mid d}\omega_N(p).
\end{align*}
Counting residue classes modulo $d$ in the interval $2\le n\le N-2$ gives
\begin{align*}
\#\{2\le n\le N-2:d\mid n(N-n)\}=\frac{\omega_N(d)}{d}N+O(\omega_N(d)).
\end{align*}
[/guided]
[/step]
[step:Apply the dimension-two Selberg upper-bound sieve]
Let $S_N$ be the number of integers $n$ with $2\le n\le N-2$ such that neither $n$ nor $N-n$ has an odd prime divisor below $N^{1/2}$. Applying the standard Selberg upper-bound sieve with level $D=N^{1/2}$ to the local density $\omega_N$ and the remainder terms $O(\omega_N(d))$ gives
\begin{align*}
S_N\ll N\prod_{2<p<N^{1/2}}\left(1-\frac{\omega_N(p)}{p}\right).
\end{align*}
The Mertens product computation for these local factors gives
\begin{align*}
\prod_{2<p<N^{1/2}}\left(1-\frac{\omega_N(p)}{p}\right)\ll \frac{\mathfrak S(N)}{(\log N)^2}.
\end{align*}
Therefore
\begin{align*}
S_N\ll \mathfrak S(N)\frac{N}{(\log N)^2}.
\end{align*}
[guided]
The local-density computation supplies exactly the hypotheses needed for the dimension-two Selberg upper-bound sieve: a main term $\omega_N(d)N/d$ for squarefree $d$ and a divisor-bounded remainder $O(\omega_N(d))$. Taking the sieve level $D=N^{1/2}$ and the same sifting cutoff gives
\begin{align*}
S_N\ll N\prod_{2<p<N^{1/2}}\left(1-\frac{\omega_N(p)}{p}\right).
\end{align*}
For primes $p\nmid N$, the factor is $1-2/p$; for primes $p\mid N$, it is $1-1/p$. The standard Mertens evaluation of this product is
\begin{align*}
\prod_{2<p<N^{1/2}}\left(1-\frac{\omega_N(p)}{p}\right)\ll \frac{\mathfrak S(N)}{(\log N)^2}.
\end{align*}
Substitution yields
\begin{align*}
S_N\ll \mathfrak S(N)\frac{N}{(\log N)^2}.
\end{align*}
[/guided]
[/step]
[step:Compare Goldbach representations with the sifted set]
If $p_1+p_2=N$ and both $p_1>N^{1/2}$ and $p_2>N^{1/2}$, then $p_1$ and $p_2$ have no odd prime divisors below $N^{1/2}$, so the parameter $n=p_1$ is counted by $S_N$. The remaining representations have $p_1\le N^{1/2}$ or $p_2\le N^{1/2}$, so there are $O(N^{1/2})$ of them. Thus
\begin{align*}
r(N)\le S_N+O(N^{1/2})
\end{align*}
for the ordered convention up to changing the absolute constant.
[guided]
A Goldbach representation corresponds to the parameter $n=p_1$, with $N-n=p_2$. If both primes are larger than $N^{1/2}$, neither can have an odd prime divisor below $N^{1/2}$, because a prime has only itself as a positive prime divisor. Hence this parameter is counted by $S_N$.
The only exceptions have $p_1\le N^{1/2}$ or $p_2\le N^{1/2}$. There are $O(N^{1/2})$ possible values in these endpoint ranges. Therefore
\begin{align*}
r(N)\le S_N+O(N^{1/2})
\end{align*}
after absorbing the ordered-pair convention into the absolute constant.
[/guided]
[/step]
[step:Absorb the endpoint term]
The Euler product defining $\mathfrak S(N)$ is bounded below by a positive absolute constant for even $N$, because the base product over $p>2$ converges to a positive number and every factor $(p-1)/(p-2)$ with $p\mid N$ and $p>2$ is greater than $1$. Hence
\begin{align*}
N^{1/2}=O\left(\mathfrak S(N)\frac{N}{(\log N)^2}\right)
\end{align*}
for sufficiently large even $N$. Combining this with the sifted-set estimate gives
\begin{align*}
r(N)\ll \mathfrak S(N)\frac{N}{(\log N)^2}.
\end{align*}
[guided]
For even $N$, the singular series satisfies $\mathfrak S(N)\ge c_0$ for some absolute constant $c_0>0$: the base Euler product is positive, and each extra factor from an odd prime divisor of $N$ is greater than $1$. Since $N^{1/2}=O(N(\log N)^{-2})$, we get
\begin{align*}
N^{1/2}=O\left(\mathfrak S(N)\frac{N}{(\log N)^2}\right).
\end{align*}
Together with
\begin{align*}
S_N\ll \mathfrak S(N)\frac{N}{(\log N)^2}
\end{align*}
and
\begin{align*}
r(N)\le S_N+O(N^{1/2}),
\end{align*}
this proves
\begin{align*}
r(N)\ll \mathfrak S(N)\frac{N}{(\log N)^2}.
\end{align*}
[/guided]
[/step]
