I have recently become aware of the prime number theorem for arithmetic progressions, but none of my textbooks seem to say anything about it. In addition, when I try to conduct an online search, the prime number theorem keeps coming up.
So, what I've been able to tell so far is:
If we let $a, b$ be positive integers such that $(a,b) = 1$, then de la Vallée Poussin's theorem for arithmetic progressions asserts that for the arithmetic sequence $\lbrace bn + a \rbrace_{n = 0}^{\infty}$,
$$ \pi_{b, a}(x) \sim \frac{1}{\phi(b)} \frac{x}{\ln x}, \hskip 5pt x \to \infty $$
where $\phi$ is Euler's totient function.
(I have used $\ln x$ in lieu of $\log x$ because it is mu understanding that it's the natural log which is meant when $\log x$ is used. I hope this is O.K.)
Now, considering that every prime is either of the form $4n + 1$ or $4n + 3$, and $\phi(4) = 2$, in both cases, we have
$$ \pi_{4, 1}(x) = \pi_{4, 3}(x) \sim \frac{1}{\phi(4)} \frac{x}{\ln x} = \frac{1}{2} \frac{x}{\ln x} $$
If I haven't made any mistakes in interpretation up to this point, my question is
(a) From the above, may I conclude that the total number of primes of the form $4n + 1$ and the total number of primes of the form $4n + 3$ are approximately equal?
and, if that is true,
(b) May I also conclude that if I pick an arbitrary prime $p$ from the set of primes, the probability that it is of the form $4n + 1$ (or $4n +3$) is approximately $1/2$? Can I say, exactly $1/2$?
and finally,
(c) May I say that for any open (or closed) interval I of real numbers, the expectation is that approximately $50\%$ of the primes contained therein are of the form $4k + 1$ and approximately $50\%$ contained therein are of the form $4k + 3$?
Thank you. (My understanding of number theory is very elementary and so I would be grateful if any responses would take that into consideration.)
Best Answer
You're generally correct.
Statements (a) and (b) are slightly unclear, given that it is impossible to pick an arbitrary number from an infinite set. This is traditionally formalized using natural density: instead of picking a "random" prime $p$ from the set of all primes (which we can't do) we pick one from the set of primes between $1$ and $N$, where $N$ is big. As $N$ gets larger and larger, this in some sense "approaches" picking an arbitrary number from an infinite set, and, by de la Vallée-Poussin's Prime Number Theorem the probability that your prime is of the form $4n+1$ approaches 50%. Your statements (a) and (b) are natural ways of stating this result, though technically mathematically imprecise.
Statement (c) is more mathematically precise, but, as abiessu noted in the comments, there is likely to be more variance if the interval only contains a few prime numbers.
However, there is an interesting note: Chebyshev's bias. Since the probability that a prime randomly chosen in the interval $[1,N]$ is of the form $4n+1$ tends to 50% as $N\to\infty$, you would expect that, for about 50% of $N$ (using the same natural density definition of "50%"), there are more $4n+1$ primes in $[1,N]$ than $4n+3$ primes, or vice versa. However, this seems to be not the case (you can read the Wikipedia article for more information, including numerical results).
It's important to note that something like de la Vallée-Poussin's Prime Number Theorem doesn't imply that $4n+1$ primes should "win out" half the time: say the first $500$ primes were all of the form $4n+1$ and then after that they alternated. After $2k+500$ primes, $k$ of them would be $4n+3$ and $k+500$ of them would be $4n+1$, and as
$$\frac{k}{2k+500}\to \frac{1}{2},\ \ \frac{k+500}{2k+500}\to \frac{1}{2},$$
the natural density statement of de la Vallée-Poussin would still hold. However, $k+500>k$, so $4n+1$ primes would win the race all the time. This is obviously not how the prime numbers are distributed, but serves as an example to show that not all sequences with natural density $\frac{1}{2}$ are "perfectly distributed" in every sense.