Let us introduce a multiplicative function supported on squarefrees by $\alpha(p)=\mathbf{1}_{p \equiv \pm 1 \bmod 8}$.
We are interested in
$$\sum_{\substack{n \le x\\ n \equiv 1 \bmod 8\\ n \text{ has }\ge 2 \text{ prime factors}}} \alpha(n) \gg \frac{x}{\sqrt{\log x}}.$$
The primes have density $1/\log x = o(1/\sqrt{\log x})$, so we may omit the condition of having at least two primes factors.
Let us, for the moment, forget about the condition $n \equiv 1 \bmod 8$. The sum $\sum_{n \le x} \alpha(n)$ can be estimated by Wirsing's Theorem (the main theorem of 'Das asymptotische Verhalten von Summen über multiplikative Funktionen. II' (Acta Math. Acad. Sci. Hungar. 18 (1967), 411–467), giving
$$ \sum_{n \le x} \alpha(n) \sim C\frac{x}{\log x} \prod_{p\le x} \left( 1+ \frac{\alpha(p)}{p}\right) = C\frac{x}{\log x} \prod_{\substack{p\equiv \pm 1 \bmod 8\\ p \le x}} \left( 1+ \frac{1}{p}\right) \sim D \frac{x}{\sqrt{\log x}}$$
for some positive constants $C,D$ depending on $\alpha$. An arithmetic input to these results is $\sum_{p \equiv a \bmod q, \, p \le x} 1/p = \log \log x/\phi(q) + O(1)$ with $q=8$.
A few remarks:
- One alternative approach is the L-S-D (Landau-Selberg-Delange) method, as suggested by Lucia. Its advantage is that it gives an asymptotic expansion and not only a main term. Tenenbaum's book has good exposition on that. A guided exercise about Landau's original proof, mentioned by Lucia as well, appears in page 187 of Montgomery and Vaughan's book. Granville and Koukoulopoulos have a paper that gives results similar to the L-S-D method, but avoiding complex analysis ('Beyond the LSD method for the partial sums of multiplicative functions
').
- A second alternative approach, which is elementary but very sophisticated, is Iwaniec's half-dimensional sieve.
- There might be a simple (and non sieve-theoretic) elementary way to study $\sum_{n \le x} \alpha(n)$; see pages 182-184 of Selberg's Collected Papers, Vol II. There he derives quickly and elementarily a variant of Landau's result on sums of two squares.
- If you are only interested in upper bounds, you could use a special case of Shiu's Theorem (`A Brun-Titschmarsh theorem for multiplicative functions'), where again the input will be $\sum_{p \le x, \, p \equiv a \bmod q} 1/p$.
Now, let us introduce back the condition $n \equiv 1 \bmod 8$. Let us define a multiplicative Möbius-like function:$$ \widetilde{\mu}(p) = \begin{cases} -1 & p \equiv -1 \bmod 8, \\ 1 & p \equiv 1 \bmod 8, \\ 0 & \text{otherwise.}\end{cases}$$
If $n$ is in the support of $\alpha$, $n \equiv (-1)^{\#\{ p \mid n : p \equiv -1 \bmod 8\}} \bmod 8$, so the condition we need to add is equivalent to $\widetilde{\mu}(n) = 1$. In other words, you are interested in evaluating $$\sum_{n \le x} \alpha(n) \frac{1+\widetilde{\mu}(n)}{2}$$
asymptotically. Since our Möbius-like function is oscillating, we would expect cancellation in $$\sum_{n \le x} \widetilde{\mu}(n)\alpha(n)$$
over the original sum. Such cancellation can be exhibited again by the L-S-D method. However, in the same paper of Wirsing mentioned above, there is a theorem (Satz 1.2.2) guarantying $$\sum_{n \le x} \widetilde{\mu}(n)\alpha(n) = o\left( \sum_{n \le x} \alpha(n)\right).$$
More generally, $\sum_{n \le x} \beta(n)/\sum_{n \le x} \alpha(n)$ can be asymptotically estimated if $\beta$ is dominated by $\alpha$ and $\alpha$ is `nice' enough.
Best Answer
The intuition is that $\log_{c\log N}N=\frac{\log N}{\log(c\log N)}=\frac{\log N}{\log c+\log\log N}$ which is asymptotically $\frac{\log N}{\log\log N}$. For this to work, we need that the kth prime for $k=\frac{\log N}{\log(c\log N)}+\pi(c\log N)\approx\frac{\log N}{\log\log N}$ to be 'close' to $c\log N$ -- actually, any constant multiple of $\log N$ will do.
$p_k\approx\frac{\log N}{\log\log N}\log\frac{\log N}{\log\log N}\approx\log N$, so $\log_{p_k}N\approx\frac{\log N}{\log\log N}$, as desired. This could probably made explicit with Rosser's theorem and/or Dusart's various bounds on $p_n$ and $\pi(n)$.