Does there exist positive rational $s$ for which the Riemann Zeta function $\zeta(s) \in N$ or equivalently, does there exist finite positive integers $\ell,m$ and $n$ such that $$\zeta\left(1+\dfrac{\ell}{m}\right) = n$$

My progress: Using the method described in my answer below, I am running a computer program, I have been able to show that if there is a solution then $l > 2.21\times 10^4$.

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    $\begingroup$ I would expect that $\zeta(r)$ is irrational for all rationals $r > 1$, but to prove this may be beyond the current state of the art. We don't even know that $\zeta(5)$ is irrational. $\endgroup$ – Robert Israel Aug 19 '18 at 5:54
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    $\begingroup$ If $\zeta$ denotes the Riemann zeta function, please include it in your question. $\endgroup$ – Klangen Aug 22 '18 at 13:07

I misread $l$ as $1$, but in any case, as a partial result, here's a resolution for the case $l=1$.

Fix $s\in\mathbb{R}$, with $s > 1$.

On the interval $(0,\infty)$, let $f(x)={\small{{\displaystyle{\frac{1}{x^{\large{s}}}}}}}$.

It's easily verified that $ {\displaystyle{ \int_{1}^\infty \!f(x)\,dx = {\small{\frac{1}{s-1}}} }} $.

Consider the infinite series $ {\displaystyle{ \sum_{k=1}^\infty \frac{1}{k^s} }} $.

Since $f$ is positive, continuous, and strictly decreasing, we get \begin{align*} \int_{1}^\infty \!f(x)\,dx < \;&\sum_{k=1}^\infty \frac{1}{k^s} < 1+\int_{1}^\infty \!f(x)\,dx\\[4pt] \implies\;{\small{\frac{1}{s-1}}} < \;&\sum_{k=1}^\infty \frac{1}{k^s} < 1+{\small{\frac{1}{s-1}}}\\[4pt] \end{align*} If $m$ is a positive integer, then letting $s=1+{\large{\frac{1}{m}}}$, we have ${\large{\frac{1}{s-1}}}=m$, hence \begin{align*} {\small{\frac{1}{s-1}}} < \;&\sum_{k=1}^\infty \frac{1}{k^s} < \;1+{\small{\frac{1}{s-1}}}\\[4pt] \implies\;m < \;\,&\zeta\bigl(1+{\small{\frac{1}{m}}}\bigr) < \;m + 1\\[4pt] \end{align*} so $\zeta\bigl(1+{\large{\frac{1}{m}}}\bigr)$ is not an integer.

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    $\begingroup$ Thanks. In fact I have a slightly stronger result can be one of the approach to tackle this problem. Using the Stieltjes series expansion of thte Riemann Zeta function we can show that for $l = 1$, the fractional part of $\zeta(1+1/m)$ starts from $\pi^2/2 - 1 = 0.6449341$ for $m = 1$ and strictly decreases with $m$ and approaches the limiting value of $1-\gamma \sim 0.422785$ where $\gamma$ is the Euler-Mascheroni constant. Hence $\pi^2/2 - 1 \le (\zeta(1+1/m)) \le \gamma$. I have been trying to see if this method can be generalized for the case $l > 1$. $\endgroup$ – Nilotpal Kanti Sinha Aug 19 '18 at 10:52
  • $\begingroup$ Yes, I had that result, but it doesn't go anywhere for $l > 1$. $\endgroup$ – quasi Aug 19 '18 at 11:02
  • $\begingroup$ Can you resolve the problem for any other value of $l$, other than $l=1$? For example, can you resolve the case $l=2$? $\endgroup$ – quasi Aug 19 '18 at 11:04
  • $\begingroup$ Note that for positive integers $l,m$, the expression $1+l/m$ can take any rational value greater than $1$, so your question can be recast as asking if there exists a rational number $s > 1$ such that $\zeta(s)$ is an integer. My sense (seconding Robert Israel's comment) is that an answer to that question is out of range of current knowledge. $\endgroup$ – quasi Aug 19 '18 at 11:09
  • $\begingroup$ I think that the case for integer should be easier than the case for deciding rationality. I will outline my approach below since it will be too long for a comment. $\endgroup$ – Nilotpal Kanti Sinha Aug 20 '18 at 6:36

Can you resolve the problem for any other value of l, other than l=1? For example, can you resolve the case l=2?

Yes and in fact I can show that $l \ge 2.2*10^4$. Here is the outline of my approach which I am posting as an answer since it is too long to be a comment.

Step 1: The first step was to derive the following result

For every integer $n \ge 2$ there exists a positive real $c_n$ such that ${\displaystyle{ \zeta\Big(1+\frac{1}{n-1+c_n}\Big) = n. }}$

The first few terms of the asymptotic expansion of $c_n$ in terms of $n$ and the Stieltjes constants $\gamma_i$ are

$$ c_n = 1-\gamma_0 + \frac{\gamma_1}{n-1} + \frac{\gamma_2 + \gamma_1 - \gamma_0 \gamma_1}{(n-1)^2} + O\Big(\frac{1}{n^3}\Big) $$

Step 2: I computed the first few values of $c_n$ but I did not use the above result. Instead I used the following recurrence formula.

Let $\alpha_0$ be any positive real and ${\displaystyle{ \alpha_{r+1} = n + \alpha_r - \zeta\Big(1+\frac{1}{n -1 + \alpha_r}\Big); }}$ then ${\displaystyle{ \lim_{r \to \infty}\alpha_r = c_n}}$.

Using this we obtained $$ c_2 \approx 0.3724062 $$ $$ c_3 \approx 0.3932265 $$ $$ \ldots $$ $$ c_{12} \approx 0.4164435 $$

Step 3: Show that $l \ge 5$

Let ${\displaystyle{ \zeta\Big(1+\frac{l}{m}\Big) \in N}}$ and let $m = lk+d$ where $\gcd(l,d) = 1$ and $1 \le d < l$.

Clearly, $c_2 \le c_n < 1-\gamma_0$ or $0.3724062 \le c_n < 0.422785$. Hence we must have ${\displaystyle{ 0.3724062 \le \frac{d}{l} < 0.422785}}$. The fraction with the smallest value of $l$ satisfying this condition is ${\displaystyle{\frac{2}{5} }}$ hence $l \ge 5$.

Extending the same approach I am able to show that $l > 2.2*10^4$.

Problems with this approach:

With this approach and with powerful computing, we can prove results like if ${\displaystyle{ \zeta\Big(1+\frac{l}{m}\Big) \in N }}$ then $l$ must be greater than some large positive integer but I don't see how this approach will solve the general problem.

Source code:

# Program with maximum n
from time import time
from mpmath import mp

start_time = time()
a = 1
a_end = 10^5
n_max = 2267
# Maximum n is at:', 4468, 1889, 2267

while(a < a_end + 1):
    b = 1 + floor(0.372406215900714*a)
    while(b <= floor((1 - euler_gamma)*a)):
        if(gcd(b,a) == 1):
            n = 2
            found = 1
            while (found == 1):
                i = 1
                r = 50
                c_n = c_n1 = N((1 - euler_gamma), digits = 100)
                while (i <= r):
                    c_n  = N(n + c_n - zeta(1 + 1/(n - 1 + c_n)), digits = 10)
                    c_n1 = N(n + 1 + c_n1 - zeta(1 + 1/(n + 1 - 1 + c_n1)), digits = 10)
                    i = i + 1
                test = N(b/a, digits = 100)
                if(c_n < test):
                    if(test < c_n1):
                        found = found - 1
                        # print(b, a, n, c_n, b/a.n(), c_n1)
                        if (n > n_max):
                            n_max = n
                            print("Maximum n is at:", a, b, n_max)
                        b = b + 1
                        if(b > floor((1 - euler_gamma)*a)):
                            found = found - 1
                            n = n + 1
                    n = n + 1
                    n = n + 1
            if(found == 1):
                found = found - 1
                print("Solutions may be found for", a, b/a, c_n, b/a.n(), c_n1)
                b = b + 1
            b = b + 1
    if(a%10^1 == 0):
        print("Checked till", a, "Duration", floor(time() - start_time))
    a = a + 1
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    $\begingroup$ Nice work! Seems like a lot of progress. Is this your own problem? $\endgroup$ – quasi Aug 20 '18 at 9:37
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    $\begingroup$ Thanks. Yes its my own problem that I worked way back in 2005 and then it got lost among other things. Revisiting it now after 13 years. $\endgroup$ – Nilotpal Kanti Sinha Aug 20 '18 at 9:52
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    $\begingroup$ This certainly seems worth publishing. Here is one possibility: tandfonline.com/loi/uexm20 $\endgroup$ – marty cohen Oct 4 '18 at 5:47

Whilst not a solution, I thought it might be interesting to see What happens if we use the functional equation:


$\zeta(s)$ is known to be rational at the negative integers,


You actually get $\zeta(-a)=0$ for $a$ even due to the trivial zeros.

But it appears that when you combine this with the functional equation, then in the limit, you get something non-zero and irrational, so that doesn't even give you an integer.

On the other hand if $m/n=2k-1$ is odd then we have


by Euler's formula, which is "almost" rational except for the $\pi$ factor. Bugger. So no chance of being integral.

If you modify your problem slightly, then using the above you can produce positive integers, but that's no fun.


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