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Assuming Riemann Hypothesis Hardy showed primes $3\bmod 4$ are more common than primes $1\bmod 4$ https://en.wikipedia.org/wiki/Riemann_hypothesis#Consequences_of_the_generalized_Riemann_hypothesis.

Is there any reason to believe there are asymptotically equal number of consecutive primes that are $1\bmod 4$ and $3\bmod 4$ with gap $4$?

Is there a possibility that there are only finitely many primes of $1\bmod 4$ form with gap $4$ while there are infinite of them which are $3\bmod 4$?

What is the intuitive reason why primes $3\bmod 4$ are more frequent?

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  • $\begingroup$ Why should there be? The Hardy-Littlewood conjecture asserts that for all admissible $k$-tuples $(a_1, \cdots, a_k)$, there are infinitely many $n$ such that $(n+a_1, \cdots, n+a_k)$ is a prime $k$-tuple. Since $(n, n+4)$ does not cover all congruence classes of any prime, it should therefore be simultaneously prime infinitely often. There are absolutely no obstructions to $(n,n+4)$ being simultaneously congruent to $3 \pmod{4}$ and prime. $\endgroup$
    – Stanley Yao Xiao
    Commented Sep 5, 2015 at 14:11
  • $\begingroup$ I think you misunderstood my problem. I ask whether among primes with gap $4$, is there a possibility that primes $1\mod 4$ occur only finitely many times while $3\bmod 4$ occur infinitely many times. $\endgroup$
    – user76479
    Commented Sep 5, 2015 at 14:13
  • $\begingroup$ Why do you expect that to be the case? It is an extremely strong statement to assert that any prime pattern with no obvious obstructions occur only finitely many times, as that would violate strongly everything we expect to be true about primes (but usually cannot prove). $\endgroup$
    – Stanley Yao Xiao
    Commented Sep 5, 2015 at 14:14
  • $\begingroup$ Because it is known a weaker version of Knapowski and Turán conjecture is true where logaithmic density of the numbers $x$ for which $π(x; 4, 3) > π(x; 4, 1)$ is high. $\endgroup$
    – user76479
    Commented Sep 5, 2015 at 14:17
  • $\begingroup$ That is true, but that still doesn't justify why you think such pairs should be finite. In fact, the natural density of pairs of primes of the form $(p, p+4)$ with $p \equiv 3 \pmod{4}$ and $p \equiv 1 \pmod{4}$ should be the same. $\endgroup$
    – Stanley Yao Xiao
    Commented Sep 5, 2015 at 14:20

2 Answers 2

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The answer ought to be 'no', that one should not expect there to only be finitely many prime pairs of the form $(4k+1, 4k+5)$ while expecting infinitely many pairs of the form $(4k+3, 4k+7)$. While we cannot prove this, it is known that weaker version of this is true.

Consider a $k$-tuple of linear functions $(L_1(n), \cdots, L_k(n)) = (a_1 n + b_1, \cdots, a_k n + b_k)$ with the property that there is no prime $p$ which divides $\prod_{i=1}^k L_i(n)$ for all integers $n$. Then it is known, from the seminal work of James Maynard (see here: http://annals.math.princeton.edu/2015/181-1/p07. This was independently discovered by Terry Tao), that there exist infinitely many $n$ such that a proportion of $(1/4 + o(1))\log k$ of the values $(L_1(n), \cdots, L_k(n))$ are prime. It is widely believed that the statement can be strengthened to assert that there are infinitely many $n$ (with a predicted density even) such that all terms in $(L_1(n), \cdots, L_k(n))$ are simultaneously prime.

If we apply this to the tuple $(4n+1, 4n+5)$ and $(4n+3, 4n+7)$ which satisfy the hypothesis above, then the assertion is true. Notice that the theorem of Maynard (and Tao) does not require any additional hypotheses on the linear functions $L_j(n)$ except what is absolutely needed in order to make the problem non-trivial.

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  • $\begingroup$ So for every $2^n$, we expect equal number of primes mod $2^n$ with gap $2k$ of form $a\bmod 2^n,a+2k\bmod 2^n$ where $a\in\{1,3,\dots,2^n-1-2k\}$ and $k\in\{1,2,\dots,2^{n-1}-2,2^{n-1}-1\}$? $\endgroup$
    – user76479
    Commented Sep 5, 2015 at 14:54
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Since there are several questions posed, here is my go at an intuitive explanation of Chebyshev's bias. Consider the two arithmetic progressions $4 n + 1$ and $4 n + 3$:

\begin{matrix} 4n+1& 4n+3\\ \hline 1 & 3\\ 5 & 7\\ 9 & 11 \\ \vdots & \vdots \end{matrix}

Now let's pick any two numbers to generate a new composite: There are three possibilities:

1) A product of two elements from column 1 ends up in column 1.

2) A product of one element from column 1 and one from column 2 ends up in column 2.

3) A product of two elements from column 2 ends up in column 1.

So, on average there are more composites pushed into column 1 and we should therefore expect less primes in this column than in column 2.

In general, consider the arithmetic progressions $dn + a_i$, where $a_i$ and $d$ are coprime, $1\leq i \leq \phi(d)$, and $\phi(d)$ is the Euler totient function. Then we can set up the matrix $(dn + a_i)(dn + a_j) \mod d$, telling us in what arithmetic progression a composite ends up. In the case above we get the matrix
$$ \left( \begin{array}{cc} 1 & 2 \\ . & 1 \end{array} \right), $$ so we see directly that the ratio is 2:1 in favour of sending composites to $4n+1$.

Another example is the case of $5n + a_i$. The resulting matrix is then $$ \left( \begin{array}{cccc} 1 & 2 & 3 & 4 \\ . & 4 & 1 & 3 \\ . & . & 4 & 2 \\ . & . & . & 1 \\ \end{array} \right). $$ Here we see that more composites are sent to the progressions $5n+1$, and $5n+4$, so we should expect $5n+2$, and $5n+3$ to have more primes. The ratio this time is 3:2:2:3, in the order of $a_i$. The larger $\phi(d)$ is, the closer the ratios will be to 1, and in the infinity limit of $\phi(d)$ we should not be able to separate out a winner of any prime race.

None of this give any proof of anything, but as a starting point to think about this problem, i quite like this approach. For more meaty stuff, one should check out the article Chebyshev's bias

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