Vtyjkyuk

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

I have been trying to solve this question and have made some progress. Computationally I can show that such an $l$ must be greater than $41$ but obviously this approach won't answer the question. Any insights on how to approach this?

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 $l ge 41$. 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 $l ge 41$.

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.