# Does there exist positive rational $s$ for which $\zeta(s)$ is a positive integer?

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 $$\zeta\left(1+\dfrac{\ell}{m}\right) = n$$

Update: 4-Apr-2021: The following code verifies that there is no solution $$l \le N$$ in $$O(N^2)$$ time complexity. Using this code and the method described in my answer below, I have been able to show that if there is a solution then $$l > 3\times 10^4$$.

a          = 1
n_max      = 2
count_max  = 0
r          = 15
max_digits = 60
eg         = N((1 - euler_gamma), digits = max_digits)
n_start    = 2
stop       = False

while stop == False:
b = 1 + floor(0.372406215900714*a)
b_max = floor(eg*a)

while(b <= b_max):
if(gcd(b,a) == 1):
test    = N(b/a, digits = max_digits)
n       = max(2,floor(0.0728158454836767248605863758/(eg - test)))
n_start = n
c_prev  = 0
i       = 1

while (i <= r):
c_prev = N(n+c_prev - zeta(1+1/(n-1+c_prev)),digits = max_digits)
i = i + 1

found = 1

while (found == 1):
c_n  = c_prev
c_n1 = eg
i    = 1

while i <= r:
c_n1 = N(n + 1 + c_n1 - zeta(1 + 1/(n + c_n1)), digits = max_digits)
i = i + 1
c_prev = c_n1

if test < c_n1:
found = 0

if (n > n_max):
n_max = n
count_max = count_max + 1
a_max = a
print("Maximum n is at:", a, b, n_max, count_max)
n = n + 1
b = b + 1
print('There is no solution for', a)
a = a + 1

• 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. – Robert Israel Aug 19 '18 at 5:54
• If $\zeta$ denotes the Riemann zeta function, please include it in your question. – 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.

• 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 $l > 1$. – Nilotpal Sinha Aug 19 '18 at 10:52
• Yes, I had that result, but it doesn't go anywhere for $l > 1$. – quasi Aug 19 '18 at 11:02
• Can you resolve the problem for any other value of $l$, other than $l=1$? For example, can you resolve the case $l=2$? – quasi Aug 19 '18 at 11:04
• 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. – quasi Aug 19 '18 at 11:09
• 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. – Nilotpal 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 there is no solution for $$l \le 3*10^4$$. 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 real $$x \ge 1$$ there exists a positive real $$c_x$$ such that $${\displaystyle{ \zeta\Big(1+\frac{1}{x-1+c_x}\Big) = x. }}$$

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

$$c_x = 1-\gamma_0 + \frac{\gamma_1}{x-1} + \frac{\gamma_2 + \gamma_1 - \gamma_0 \gamma_1}{(x-1)^2} + \frac{\gamma_2 +2\gamma_2 - 2\gamma_2 \gamma_0 + \gamma_1 - 2\gamma_1 \gamma_0 + \gamma_1 \gamma_0^2 - \gamma_1^2}{(x-1)^3} + O\Big(\frac{1}{x^4}\Big)$$

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 this approach using numerical computations, I am able to show that there is no solution for $$l < 3*10^4$$.

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.

• Nice work! Seems like a lot of progress. Is this your own problem? – quasi Aug 20 '18 at 9:37
• 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. – Nilotpal Sinha Aug 20 '18 at 9:52
• This certainly seems worth publishing. Here is one possibility: tandfonline.com/loi/uexm20 – 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:

$$\zeta\left(1+\frac{m}{n}\right)=2^{\left(1+\frac{m}{n}\right)}\pi^{\frac{m}{n}}\sin\left(\frac{\pi}{2}\left(1+\frac{m}{n}\right)\right)\Gamma\left(-\frac{m}{n}\right)\zeta\left(-\frac{m}{n}\right).$$

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

$$\zeta(-a)=(-1)^a\frac{B_{a+1}}{a+1}.$$

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

$$\zeta\left(2k\right)=\frac{(-1)^{k+1}B_{2k}(2\pi)^{2k}}{2(2k)!},$$

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.