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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 $$zetaleft(1+dfracellmright) = n$$

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 $l > 2.21times 10^4$.

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

Fix $sinmathbbR$, with $s > 1$.

On the interval $(0,infty)$, let
$f(x)=smalldisplaystylefrac1x^larges$.

It's easily verified that
$
displaystyle
int_1^infty !f(x),dx
=
smallfrac1s-1

$.

Consider the infinite series
$
displaystyle
sum_k=1^infty frac1k^s

$.

Since $f$ is positive, continuous, and strictly decreasing, we get
beginalign*
int_1^infty !f(x),dx < ;&sum_k=1^infty frac1k^s < 1+int_1^infty !f(x),dx\[4pt]
implies;smallfrac1s-1 < ;&sum_k=1^infty frac1k^s < 1+smallfrac1s-1\[4pt]
endalign*
If $m$ is a positive integer, then letting $s=1+largefrac1m$, we have $largefrac1s-1=m$, hence
beginalign*
smallfrac1s-1 < ;&sum_k=1^infty frac1k^s < ;1+smallfrac1s-1\[4pt]
implies;m < ;,&zetabigl(1+smallfrac1mbigr) < ;m + 1\[4pt]
endalign*
so $zetabigl(1+largefrac1mbigr)$ is not an integer.

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 zetaBig(1+frac1n-1+c_nBig) = n. $

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

$$ c_n = 1-gamma_0 + fracgamma_1n-1
+ fracgamma_2 + gamma_1 - gamma_0 gamma_1(n-1)^2 + OBig(frac1n^3Big) $$

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 - zetaBig(1+frac1n -1 + alpha_rBig); $ then $displaystyle lim_r to inftyalpha_r = c_n$.

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 $displaystyle zetaBig(1+fraclmBig) in N$ 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 $displaystyle 0.3724062 le fracdl < 0.422785$. The fraction with the smallest value of $l$ satisfying this condition is $displaystylefrac25 $ 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 zetaBig(1+fraclmBig) 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
 else:
 n = n + 1
 n = n + 1
 else:
 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
 else:
 b = b + 1
 if(a%10^1 == 0):
 print("Checked till", a, "Duration", floor(time() - start_time))
 a = a + 1

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

$$zetaleft(1+fracmnright)=2^left(1+fracmnright)pi^fracmnsinleft(fracpi2left(1+fracmnright)right)Gammaleft(-fracmnright)zetaleft(-fracmnright).$$

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

$$zeta(-a)=(-1)^afracB_a+1a+1.$$

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

$$zetaleft(2kright)=frac(-1)^k+1B_2k(2pi)^2k2(2k)!,$$

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.