TITLE: Elementary lower bounds for the number of primes in arithmetic progressions
QUESTION [15 upvotes]: Some version of the Prime Number Theorem provides the asymptotic behavior of the number of primes in arithmetic progression $qn+a$ with $(q,a)=1$, $n \ge 1$. I was wondering there are Chebyshev-type arguments (using the binomial coefficients or variants thereof) that establish the existence of at least $$c x/\log x$$ primes in such arithmetic progressions up to $x$ for some even small $c>0$ and some values of $a, q>1$. For instance, is there such an argument for primes of the form $ 4k \pm 1$? This is mostly a curiosity triggered by thinking about a number theory course I will teach soon.

REPLY [4 votes]: Here is another, more Chebyshev like, approach. It is possible that this is the proof Terry was sketching and I just didn't understand it. I think this should generalize to AP's for any modulus, but I'll stick to $4$. Let $\chi$ be the quadratic character modulo $4$. To prove PNT in AP's mod 4, we must show
$$\sum_{n \leq N} \Lambda(n) = N + o(N),\ \sum_{n \leq N} \chi(n) \Lambda(n) = o(N).$$
Chebyshev style arguments prove bounds of the form $a N < \sum_{n \leq N} \Lambda(n) < b N$; I will analogously prove bounds of the form $\left| \sum_{n \leq N} \chi(n) \Lambda(n) \right| < cN$. This will only be interesting if I manage to get $c<1$.
For any positive real number $x$, define 
$$\langle x \rangle = \sum_{1 \leq k \leq x} \chi(k).$$
So $\langle x \rangle$ is $1$ if $\lfloor x \rfloor \equiv 1,2 \bmod 4$ and $0$ if $\lfloor x \rfloor \equiv 3,4 \bmod 4$.
Put 
$$S = \sum_{n \leq N} \chi(n) \log n.$$
Start by noticing that
$$\left| S \right| \leq \log N$$
because each term switches the sign of the partial sum.
Expand the sum as 
$$S = \sum_{n \leq N} \sum_{d|n} \chi(d) \chi(n/d) \Lambda(d) = \sum_d \chi(d) \Lambda(d) \sum_{m \leq N/d} \chi(m) = \sum_d \chi(d) \Lambda(d) \ \langle\! N/d \rangle. \quad (\ast)$$
Now, $\langle N/d \rangle$ is $1$ for $N/3 < d \leq N$, is $0$ for $N/5 < d \leq N/3$ and is between $0$ and $1$ for all $d$, so
$$\left|\sum_{N/3< d \leq N} \chi(d) \Lambda(d) \right| \leq |S| + \sum_{d \leq N/5} \Lambda(d).$$
Invoking Chebyshev's bound $\sum_{d \leq M} \Lambda(d) \leq 1.11 M$, we have 
$$\left|\sum_{N/3< d \leq N} \chi(d) \Lambda(d) \right| \leq \log N + 1.11 \times \frac{1}{5} N .$$
Summing up this equation for $N$, $N/3$, $N/9$, ..., we get
$$\left|\sum_{d \leq N} \chi(d) \Lambda(d) \right| \leq \frac{(\log N)^2}{\log 3} + 1.11 \times \frac{3}{10} N.$$
Since $0.333<1$, we win.
I think there is substantial room to improve this argument. If we sum up equation $(\ast)$ at $N$ and $N/3$, we get
$$\sum_d \chi(d) \Lambda(d) {\Big(} \langle N/d \rangle + \langle N/(3d) \rangle {\Big)} = O(\log N).$$
We have ${\Big(} \langle N/d \rangle + \langle N/(3d) \rangle {\Big)} = 1$ for $N/7 < d \leq N$, then $=0$ for $N/9 < d \leq N/7$, and is always at most $2$. So similar arguments give
$$\left| \sum_{N/7 < d \leq N} \chi(d) \Lambda(d) \right| \leq 1.11 \times \frac{2}{9} N + O(\log N)$$
and thus replace $1.11 \times \tfrac{3}{10}$ with $1.11 \times \frac{16}{63}$ in the final analysis. I think this constant could be chased much lower if we took smarter linear combinations. 
More precisely, we have $\sum_{t=1}^T \chi(t) \mu(t) \langle x/t \rangle = 1$ for $x \geq T$. So, for fixed $T$, we have
$$\sum_{t=1}^T \mu(t) \chi(t) \sum_d \chi(d) \Lambda(d) \ \langle\! N/(td) \rangle =$$
$$\sum_{N/T < d \leq N} \chi(d) \Lambda(d)  + \sum_{d \leq N/T} \chi(d) \Lambda(d) \left( \sum_{t=1}^T \mu(t) \chi(t) \langle\! N/(td) \rangle \right) = O(\log N).$$
If we could get the parenthesized quantity to be $o(T)$, independent of $N$, we would win.