In one of Soundararajan's papers, he claims without proof that it is a standard exercise to show that the number $N(X)$ of positive square-free integers $d \equiv 1 \; mod \; 8$ less than $X$, with at least two primes factors all of which is congruent to $\pm1 \; mod \; 8$ is asymptotically bounded by $\frac{X}{\sqrt{log(X)}}$ up to a constant. I am new to techniques in analytical number theory, and I find it hard to see this as a standard exercise. Based on my limited knowledge of such techniques I believe this may follow from some kind of Tauberian theorem. But, I couldn't come up with that sort of proof to the above fact. So I would like to see a standard proof of this fact.

Instead, I did some hand-wavy computations that lead me to a much stronger lower bound $\frac{X}{\sqrt{log(log))}}$ (up to a constant). For simplicity, I present my ideas to just counting those $d$ with the stated properties but with prime factors congruent to $1 \; mod \; 8$ only, the general case is a simple extension of the same ideas. There are $\frac{X^{1/r}}{8log(X^{1/r})} = \frac{rX^{1/r}}{8log(X)}$ number of primes less than $X^{1/r}$ and congruent to $1 \; mod \; 8$. So there are ${\frac{rX^{1/r}}{8log(X)} \choose r}$ number of such $d<X$. Now summing them from $r = 2$ onwards we have $N(X) = \sum_{r \geq 2} {\frac{rX^{1/r}}{8log(X)} \choose r} \gg \sum_{r \geq 2} {(\frac{rX^{1/r}}{8log(X)})}^r \frac{1}{r!} \gg X \sum_{r \geq 2} \frac{1}{{(8log(X))}^r} \frac{r^r}{r!}$. And now using Stirling's approximation, $\frac{r^r}{r!} \approx \frac{e^r}{\sqrt{2 \pi r}}$. So $N(X) \gg X \sum_{r \geq 2} {(\frac{e}{8log(X)})}^r r^{-1/2}$, approximating the sum as an integral $N(X) \gg \int_2^{\infty}a^x x^{-1/2} dx \approx \frac{\Gamma(1/2, -2 log(a))}{\sqrt{-log(a)}}\gg> \frac{1}{\sqrt{log(log(X)}}$, where $a = \frac{e}{8log(X)}$. Hence we have the result. (I use "$\gg$" above to hide constants in the inequalities)

Since my method gives a somewhat stronger bound, I am skeptical if there is some flaw in my ideas. Please feel free to comment on what I have done.

In one of Soundararajan's papers, he claims without proof that it is a standard exercise to show that the number $N(X)$ of positive square-free integers $d \equiv 1 \; \bmod \; 8$ less than $X$, with at least two primes factors all of which is congruent to $\pm1 \; \bmod \; 8$ is asymptotically bounded by $\frac{X}{\sqrt{\log(X)}}$ up to a constant. I am new to techniques in analytical number theory, and I find it hard to see this as a standard exercise. Based on my limited knowledge of such techniques I believe this may follow from some kind of Tauberian theorem. But, I couldn't come up with that sort of proof to the above fact. So I would like to see a standard proof of this fact.

Instead, I did some hand-wavy computations that lead me to a much stronger lower bound $\frac{X}{\sqrt{\log(\log X)}}$ (up to a constant). For simplicity, I present my ideas to just counting those $d$ with the stated properties but with prime factors congruent to $1 \; \bmod \; 8$ only, the general case is a simple extension of the same ideas. There are $M = M(x,r) = \frac{X^{1/r}}{8\log(X^{1/r})} = \frac{rX^{1/r}}{8\log(X)}$ number of primes less than $X^{1/r}$ and congruent to $1 \; \bmod \; 8$. So there are $\binom{M}{r}$ number of such $d<X$. Now summing them from $r = 2$ onwards we have $$N(X) = \sum_{r \geq 2} \binom{M(x,r)}{r} \gg \sum_{r \geq 2} {\left(\frac{rX^{1/r}}{8\log(X)}\right)}^r \frac{1}{r!} \gg X \sum_{r \geq 2} \frac{1}{{(8\log(X))}^r} \frac{r^r}{r!}.$$ And now using Stirling's approximation, $\frac{r^r}{r!} \approx \frac{e^r}{\sqrt{2 \pi r}}$. So $N(X) \gg X \sum_{r \geq 2} {\left(\frac{e}{8\log(X)}\right)}^r r^{-1/2}$, approximating the sum as an integral $$N(X) \gg \int_2^{\infty}a^x x^{-1/2} dx \approx \frac{\Gamma(1/2, -2 \log(a))}{\sqrt{-\log(a)}}\gg> \frac{1}{\sqrt{\log(\log X)}},$$ where $a = \frac{e}{8\log(X)}$. Hence we have the result. (I use "$\gg$" above to hide constants in the inequalities)

Since my method gives a somewhat stronger bound, I am skeptical if there is some flaw in my ideas. Please feel free to comment on what I have done.

In one of Soundararajan's papers, he claims without proof that it is a standard exercise to show that the number $N(X)$ of positive square-free integers $d \equiv 1 \; mod \; 8$ less than $X$, with at least two primes factors all of which is congruent to $\pm1 \; mod \; 8$ is asymptotically bounded by $\frac{X}{\sqrt{log(X)}}$ up to a constant. I am new to techniques in analytical number theory, and I find it hard to see this as a standard exercise. Based on my limited knowledge of such techniques I believe this may follow from some kind of Tauberian theorem. But, I couldn't come up with that sort of proof to the above fact. So I would like to see a standard proof of this fact.

Instead, I did some hand-wavy computations that lead me to a much stronger lower bound $\frac{X}{\sqrt{log(log))}}$ (up to a constant). For simplicity, I present my ideas to just counting those $d$ with the stated properties but with prime factors congruent to $1 \; mod \; 8$ only, the general case is a simple extension of the same ideas. There are $\frac{X^{1/r}}{8log(X^{1/r})} = \frac{rX^{1/r}}{8log(X)}$ number of primes less than $X^{1/r}$ and congruent to $1 \; mod \; 8$. So there are ${\frac{rX^{1/r}}{8log(X)} \choose r}$ number of such $d<X$. Now summing them from $r = 2$ onwards we have $N(X) = \sum_{r \geq 2} {\frac{rX^{1/r}}{8log(X)} \choose r} >> \sum_{r \geq 2} {(\frac{rX^{1/r}}{8log(X)})}^r \frac{1}{r!} >> X \sum_{r \geq 2} \frac{1}{{(8log(X))}^r} \frac{r^r}{r!}$. And now using Stirling's approximation, $\frac{r^r}{r!} \approx \frac{e^r}{\sqrt{2 \pi r}}$. So $N(X) >> X \sum_{r \geq 2} {(\frac{e}{8log(X)})}^r r^{-1/2}$, approximating the sum as an integral $N(X) >> \int_2^{\infty}a^x x^{-1/2} dx \approx \frac{\Gamma(1/2, -2 log(a))}{\sqrt{-log(a)}}>> \frac{1}{\sqrt{log(log(X)}}$, where $a = \frac{e}{8log(X)}$. Hence we have the result. (I use "$>>$" above to hide constants in the inequalities)

Since my method gives a somewhat stronger bound, I am skeptical if there is some flaw in my ideas. Please feel free to comment on what I have done.

In one of Soundararajan's papers, he claims without proof that it is a standard exercise to show that the number $N(X)$ of positive square-free integers $d \equiv 1 \; mod \; 8$ less than $X$, with at least two primes factors all of which is congruent to $\pm1 \; mod \; 8$ is asymptotically bounded by $\frac{X}{\sqrt{log(X)}}$ up to a constant. I am new to techniques in analytical number theory, and I find it hard to see this as a standard exercise. Based on my limited knowledge of such techniques I believe this may follow from some kind of Tauberian theorem. But, I couldn't come up with that sort of proof to the above fact. So I would like to see a standard proof of this fact.

Instead, I did some hand-wavy computations that lead me to a much stronger lower bound $\frac{X}{\sqrt{log(log))}}$ (up to a constant). For simplicity, I present my ideas to just counting those $d$ with the stated properties but with prime factors congruent to $1 \; mod \; 8$ only, the general case is a simple extension of the same ideas. There are $\frac{X^{1/r}}{8log(X^{1/r})} = \frac{rX^{1/r}}{8log(X)}$ number of primes less than $X^{1/r}$ and congruent to $1 \; mod \; 8$. So there are ${\frac{rX^{1/r}}{8log(X)} \choose r}$ number of such $d<X$. Now summing them from $r = 2$ onwards we have $N(X) = \sum_{r \geq 2} {\frac{rX^{1/r}}{8log(X)} \choose r} \gg \sum_{r \geq 2} {(\frac{rX^{1/r}}{8log(X)})}^r \frac{1}{r!} \gg X \sum_{r \geq 2} \frac{1}{{(8log(X))}^r} \frac{r^r}{r!}$. And now using Stirling's approximation, $\frac{r^r}{r!} \approx \frac{e^r}{\sqrt{2 \pi r}}$. So $N(X) \gg X \sum_{r \geq 2} {(\frac{e}{8log(X)})}^r r^{-1/2}$, approximating the sum as an integral $N(X) \gg \int_2^{\infty}a^x x^{-1/2} dx \approx \frac{\Gamma(1/2, -2 log(a))}{\sqrt{-log(a)}}\gg> \frac{1}{\sqrt{log(log(X)}}$, where $a = \frac{e}{8log(X)}$. Hence we have the result. (I use "$\gg$" above to hide constants in the inequalities)

Since my method gives a somewhat stronger bound, I am skeptical if there is some flaw in my ideas. Please feel free to comment on what I have done.