I am studying the proof of Vinogradov's Theorem from Davenport's Multiplicative Number Theory. I am stuck with the following
It writes that the formula is of little use when $N$ is even. However it mentions that the number of ways to write an even integer as sum of three primes is $O(N(\log N)^4)$. The hint is that one of the $k_i$ must be a power of $2$. I see why one of the $k_i$ is a power of $2$, but I cannot figure out why the number of ways to write an even integer as sum of $3$ power of primes is $O(N(\log N)^4)$.
Here is what I have tried.
for example, for $k_1$ we have $\log_2 N=O(\log N)$ possibilites, and the number of possibilites for $k_2$ is bounded by $$\log_2 N+\log_3 N+\log_4 N+ \dots \log_N N=\frac{\log N}{\log 3}+ \dots \frac{\log N}{\log N}=\log N\big( \sum_{i=3}^{N} \frac{1}{\log i} \big)=O(N \log N)$$. So the number of ways to express an even number is bounded by $N(\log N)^2$. Correct, Right? but Davenport writes $N(\log N)^4$ and I am unsure why?
Note that $r(N)$ is not the number of representations of $N$ as the sum of three prime powers, but a weighted version of that: $$ r(N) = \sum_{\substack{1\le k_1,k_2,k_3\le N \\ k_1+k_2+k_3=N}} \Lambda(k_1)\Lambda(k_2)\Lambda(k_3). $$ Since we are aiming for a $O$-bound, we may permute the variables any way we wish (we lose at most a factor of $3!$ by doing so). So we may bound $$ \sum_{\substack{1\le k_1,k_2,k_3\le N \\ k_1+k_2+k_3=N \\ k_1 \text{ is a power of }2 \\ k_2,k_3\text{ are prime powers}}} \Lambda(k_1)\Lambda(k_2)\Lambda(k_3) \ll (\log N)^3 \sum_{\substack{1\le k_1,k_2,k_3\le N \\ k_1+k_2+k_3=N \\ k_1 \text{ is a power of }2 \\ k_2,k_3\text{ are prime powers}}} 1. $$ Since $k_3=N-k_1-k_2$ is uniquely determined by $k_1$ and $k_2$, we see that \begin{align*} r(N) &\ll (\log N)^3 \cdot \#\{\text{powers of $2$ less than $N$}\} \cdot \#\{\text{prime powers less than $N$}\} \\ &\ll (\log N)^3 \cdot \log N \cdot \frac N{\log N} = N(\log N)^3. \end{align*} The fact that the number of prime powers less than $N$ is $\ll N/\log N$ follows from \begin{align*} \#\{\text{prime powers less than $N$}\} &= \sum_{k=1}^{(\log N)/\log 2} \pi(N^{1/k}) \\ &= \pi(N) + \sum_{k=2}^{(\log N)/\log 2} \pi(N^{1/k}) \\ &\le \pi(N) + \sum_{k=2}^{(\log N)/\log 2} \pi(\sqrt N) \ll \frac N{\log N} + \log N\cdot \sqrt N. \end{align*}