This question requires little bit of explanation of the background hence it is a bit lengthy. Note: The question was initially posted in MSE but did not get answers hence posting in MO.

For every integer $n \ge 2$ there is a unique real $1< x_n <1.73$ such that the Riemann nzeta function $\zeta(x_n) = n$. Example $$\zeta(1.72865..) = 2, \text{ } \zeta(1.41785..) = 3, \text{ } \zeta(1.29396..) = 4, ... $$ It is likely that all $x_n$ are irrational but this is an open problem. I have devised a methodology to get some partial results in this direction.

If a rational $x_n > 1$ exists then, WLOG, let $x_n = 1 + \dfrac{a_n}{b_n} = 1 + \dfrac{1}{i_n + f_n}$ where $i_n$ is the integer part of the irreducible fraction $b_n/a_n$ and $f_n$ is its fractional part. From the Stieltjes series expansion of the Riemann zeta function, we obtain

$$ f_2 \approx 0.37241 \le f_n \le f_{n+1} < 1 - \gamma \approx 0.422785 $$ where $\gamma$ is the Euler-Mascheroni constant. $f_n$ is strictly increasing follows from the fact that $\zeta(x)$ is strictly decreasing on our region of interest $x \ge x_2$.

Thus if $\zeta(1 + a_n/b_n) = n$ then we can write $b_n = ka_n + r_n$ for some integers $k$ and $r_n$ such that $\gcd(r_n,a_n) = 1$ and $0.37241 < \dfrac{r_n}{a_n} < 0.422789$.

Now for each $a_n \ge 1$ we can eliminate the possible values of $r_n$ by computer verification to increase the lower bound on $a_n$ as shown below.

Example: Proving $a_n \ge 6$ for $n \ge 2$.

For $a_n = 1,2,3,4$, there is no $r_n$ satisfying $0.37241 < \dfrac{r_n}{a_n} < 0.422789$ so these values of $a_n$ are eliminated. For $a_n = 5$ there is one possible value, $r_n = 2$ which satisfies both $\gcd(r_n,a_n) = 1$ and $0.37241 < \dfrac{r_n}{a_n} < 0.422789$. So we take $r_n = 2$ for further test. We observe that

$$f_3 \approx 0.3932265 < \frac{2}{5} < f_4 \approx 0.4018059$$

Since $\dfrac{2}{5}$ lies between two consecutive values of $f_3$ and $f_4$ is strictly increasing, we will never have a situation where $f_n = \dfrac{2}{5}$ for any $n$. Hence $r_n = 2$ is ruled out for $a_n = 5$. The integers $3$ and $4$ are witnesses for the elimination of $\dfrac{2}{5}$. With this we have exhausted all possibilities for $a_n = 5$ and so we can conclude if $\zeta(x_n)\in N$ then $a_n \ge 6$. Similarly we can eliminate $a_n = 6,7,\ldots$ and so on.

Note: As $a_n$ increases the number of possible $r_n$ increases. Each $r_n$ must be eliminated individual with witness to completely eliminate $a_n$. So after a point, $a_n$ will have multiple witnesses.

Definition: A witness for a given integer pair $(a,r), a > r$ is an integer $w_{a,r}$ such that $f_{w_{a,r}} < \dfrac{r}{a} <f_{1+w_{a,r}}$

The existence of a witness $w_{a,r}$ is therefore sufficient to prove that $\zeta\Big(1 +\dfrac{a}{x+r}\Big)$ is not integer for any integer $x > a > r$. By actual computation, I found the witness for every permissible pair $(a,r), a \le 10^4$ which implies that

If $x_n > 1$ is rational and $\zeta(x_n) \in N$ then the difference between the numerator and denominator of $x_n$ is greater than $10^4$.

Size of the witness

I noticed that most integers have small witnesses but a few of them have large witnesses. I looked at the list of integers $a$ whose largest witness for some $r$ is larger than all the witness of all the integers less than $a$. Let us call these the maximal witnesses. Given below are the first few values of the maximal witnesses:

a    Max of w_{a,r}
5    3
12   12
19   42
45   130
30   292
97   701
123  3621
518  16503
913  31689
1308 49858
1703 71984
2098 99521
2493 134724
2888 181317
3283 245888
3678 341249
4073 496221
4468 790559
4863 1598102
5258 11274164

Some of the values $a = 5,19,123,5258$ are the non-successive denominators in the continued fraction expansion of $\gamma$ but in general each of these values of $a$ are the denominators in increasing order for the best rational approximation of $\gamma$. In other words, the closer is $\dfrac{r}{a}$ to $1-\gamma$, the larger is its witness $w_{a,r}$ which makes computationally eliminating $a$ more difficult.

This finally bring me to my question.

Question: Is the problem of finding positive integer solutions $\zeta(p/q) = u$ equivalent to deciding the rationality of $\gamma$.

Related question


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.