Are there any plausible arguments for the infinity of right shifted prime numbers?

Question

By "right shifted prime numbers" I mean prime numbers of the form:

$p_r \equiv$ $ 1 $ $mod $ $6$.

$p_l \equiv$ $5$ $mod$ $6$ on the other side would be a left shifted prime number.

Since all prime numbers $p_n>3$ are either right shifted or left shifted it seems to be a generalization of a huge amount of much more specific conjectures when simply asking if there are infinitely many right shifted prime numbers at all.

So before bothering about any folow up conjectures like "infinity of Twin primes", "infinity of Mersenne primes" and such, I'd like to understand why there should be infinitely many right shifted prime numbers, while some simple thought experiments don't suggest it:

1. Any amount of right shifted prime factors always results in a right shifted composite number.
2. Any even amount of left shifted prime factors always results in a right shifted composite number.
3. While on the other hand only an odd amount of left shifted prime factors results in a left shifted composite number.

Primorial - intervals	$30<p<210$	$210<p<2310$	$2310<p<30030$	$30030<p<510510$	$510510<p<9699690$	$9699690<p<9699690*23$
All prime numbers	36	297	2905	39083	603698	11637502
Left shifted prime numbers	18	150	1462	19560	301928	5819042
Right shifted prime numbers	18	147	1443	19523	301770	5818460
The gap of right shifts	0	3	19	37	158	582

While at a first glance the ratios of Right-shifted-primes/All-primes might seem to converge to $0.5$, I highly doubt it, I'd rather expect to see an exponential increase of the right shift gap on further primorial intervals, when all the previous right shift gaps begin to really matter.

The exponential grwoth of gaps between the Mersenne-primes also highly suggests that there aren't infinitely many right shifted primes, since all Mersenne-primes are right shifted. This would then suggest that there exists some final right shifted prime number and no other prime number from there on can ever happen to be right shifted.

Answer 1

Let $\pi(r,m;x)$ denote the count of primes $p$ less than $x$ such that $p\equiv r\,\,\mathrm{mod}\,\,{m}$. It is clear that if $r$ and $m$ are not relatively prime, i.e $\gcd(r,m)>1$, then there can't be infinitely many primes in $\pi(r,m;x)$ since any number $p\equiv r\,\, \mathrm{mod}\,\,m$ factors as $p=r+mk=\gcd(r,m)(r/\gcd(r,m)+m/\gcd(r,m)k)$.

Let $\varphi(m)$ be the count of relatively prime integers less than $m$. A natural generalization of your question is to assume whether or not the primes are in a sense equally distributed across the $\varphi(m)$ possibilities, meaning that

$$\lim_{x\to\infty}\frac{\pi(r,m;x)}{\pi(x)}=\frac{1}{\varphi(m)}$$

Where $\pi(x)$ is the total number of primes less than $x$ (i.e the percentage of primes less than $x$ is $\frac{1}{\varphi(m)}$). Your question is the case $m=6$, where we note that $\varphi(6)=2$ since $1$ and $5$ are the only numbers less than $6$ that are relatively prime to $6$.

This turns out to be very difficult to prove, but it was shown to be true in 1837 and it now known as Dirichlet's Theorem for Primes in Arithmetic Progressions. Research in this direction continues however, since people have notices than the quantities $\frac{\pi(r,m;x)}{\pi(x)}-\frac{1}{\varphi(m)}$ seem to have "preferences" for being positive or negative, meaning that for a given value of $r$ we generally see slightly less primes than expected.

This phenomenon is very interesting, and I highly recommend you check it out and learn more if you are interested.

Answer 2

Why there should be infinitely many right shifted prime numbers?

Suppose there were only finitely many and consider the number $X$ which is the product of all these primes together with $6$.

Let $p$ be a prime which divides $X^2+X+1$. Then $p$ would have to be of the form $6k+5$ since all other primes divide $X$.

By Fermat's little Theorem, $X^{6k+4}=X^{p-1}\equiv 1\pmod p$.

Since $X^3-1=(X^2+X+1)(X-1)$ we also have $X^3\equiv 1\pmod p$.

Therefore $X=X^{(6k+4)-3(2k+1)}\equiv 1\pmod p$.

But then $X^2+X+1\equiv 3\pmod p$ and is not divisible by $p$ after all.

This contradiction proves that there are infinitely many right shifted primes.