Frequency of integers $x, x+2$ such that gcd$\left(x(x+2),p\right)=1$

Question

Let:

• $p\ge 5$ be a prime.
• $p\#$ be the primorial of $p$.
• $0 < x < p\#$ be an integer.
• gcd$(a,b)$ be the greatest common divisor of $a$ and $b$.

It is straight forward to show that there are $\prod\limits_{q \text{ is odd prime, }q \le p}(q-2)$ instances of $x$ where $x < p\#$ and gcd$\left(x(x+2),p\#\right)=1$:

Base Case: There are 3 such $x$ for $p=5$ which are $\{11, 17, 29\}$

Inductive Hypothesis: Assume it is true up to some prime $p \ge 5$

Inductive Case:

• Let $x_1, x_2, \dots x_n$ fulfill this hypothesis for $p$ so that each $x_i < p\#$, gcd$\left(x_i(x_i+2),p\#\right)=1,$ and $n = \prod\limits_{q\text{ is odd prime, } q \le p}(q-2)$

• Let $r$ the least prime greater than $p$.

• Each $x_i, x_i + p\#, x_i + 2p\#, \dots, x_i + (r-1)p\#$ forms a complete residue system modulo $r$.

• As such, for each $x_i$, exactly $2$ are either congruent to $r$ or $r-2$. The remaining $r-2$ will have the property that gcd$\left(x_i + up\#)(x_i + up\#+2),r\#\right)=1$ where $0 \le u \le r-1$.

• Thus, the number of $x$ that have the desired property relative $r$ is $\left(\prod\limits_{q\text{ is odd prime, }q\le p}(q-2)\right)(r-2) = \prod\limits_{q\text{ is odd prime, }q\le r}(q-2)$

Here is my question:

Let $C_p$ be the count of $x$ such that:

• $0 <x < p\#$
• gcd$\left(x(x+2),p\#\right)=1$

I am interested in seeing if there is a bound for counting the number of $x$ for each $u$ where:

• $0 \le u < C_p$
• $u\left(\dfrac{p\#}{C_p}\right) < x \le (u+1)\left(\dfrac{p\#}{C_p}\right)$

When I look at $p \le 13$, I am finding that this count is never greater than $2$.

Is it known if this always follows? Will there ever be a prime $r$ such that the count of an interval defined as above for $r$ would have a count greater than $2$?

I am suspecting that it is straight forward to prove that $2$ is the maximum count. Am I wrong?

Is there a counter example?

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Edit: Updated question to make it clear that $q > 2$.

Answer 1

Your updated original statement & proof looks correct.

Regarding your main question about the distribution of the values of $x$ where $\gcd(x(x+2),p\#) = 1$ for various primes $p$, I'll first provide an argument why there's actually very likely no upper bound on the maximum number in any single sub-range, and then give you an example where there are $3$ such values.

First, using just the first $2$ primes of $2$ and $3$, $x \equiv 1 \pmod 2$ and $x \equiv 2 \pmod 3$ combine to be $x \equiv 5 \pmod 6$. With $p = 5$, the $3$ available modulo values are $1, 2, 4$ giving that $x \equiv 11, 17, 29 \pmod{30}$. You can see there are $2$ cases where there are $3$ values within $18$ of each other, this being $11, 17, 29$ and, with wrap-around, $29, 41, 47$.

Next, with $p = 7$, the $5$ available modulo values are $1, 2, 3, 4, 6$. Note the $2$ sequences mentioned in the previous paragraph, when considered modulo $30 \times 7 = 210$, will each be repeated $7$ times. Among these $7$ times, the ones which will be excluded are where the first, second and/or third values are congruent to $0$ or $5$ modulo $7$. This can happen at the most $3 \times 2 = 6$ times, which means each sequence will occur at least $7 - 6 = 1$ times (except for the very unusual case, which doesn't happen here, where the second sequence has only $1$ which passes and it's the last one so the wrap-around goes past $210$). For larger primes $p$, this will be a larger minimum multiplier value of $p - 6$, so these sequences of $3$ values within $18$ of each other will occur more & more often. You just need to have any one of them occurring within an integral sub-range of

$$S_p = \frac{p\#}{C_p} = 2\prod_{i=2}^{n}\frac{p_i}{p_i - 2} \tag{1}\label{eq1}$$

where for $p_i$, $i$ is the prime index (e.g., $p_1 = 2, p_2 = 3, \ldots$). Note that $\frac{p_i}{p_i - 2} \gt \frac{p_i}{p_i - 1}$ and the proof in Infinite product involving primes given in the answer by Milo Brandt shows the product of $\frac{p_i}{p_i - 1}$ for $i$ to infinity has no upper limit. This increasing # of sequences and each sub-range length strongly indicates that, unless there's some sort of weird interaction preventing it, there will always eventually be at least one instance of $3$ values of $x$ occurring in a sub-range. Note you can extend this argument to show there will also very likely eventually be instances of length $4, 5, 6, \ldots$, i.e., there's no upper limit.

To confirm there's at least $2$ sub-sequences where $3$ values are $18$ apart modulo $210$, even without using wrap-around, note the set of values are $11, 17, 29, 41, 59, 71, 101, 107, 137, 149, 167, 179, 191, 197, 209$. There are actually $3$ such sequences, with them being $11$ to $29$, $179$ to $197$ and $191$ to $209$.

In particular, note the values in the second sub-sequence of $179, 191, 197$ are each the lesser of a twin prime, so none of them will be excluded until $p \ge 179$. Using \eqref{eq1} gives that $S_{37} = 34.0511\ldots$. With $u = 5$, this gives $uS_{37} = 170.255\ldots$ and $(u + 1)S_{37} = 204.306\ldots$, showing there are $3$ such $x$ in this sub-range.

Unfortunately, I haven't provided a rigorous proof there is no upper limit. I believe, considering the details of how the various values behave with each other, it at least would likely be very difficult to give a rigorous proof. Nonetheless, based on my reasoning provided here, I'm confident there's no upper bound.