*This question requires a little background explanation and is therefore a bit lengthy.* Note: The question was originally posted in MSE but received no replies, so it was posted in MO.

For every integer $ n ge 2 $ there is a unique real $ 1 <x_n <$ 1.73 so 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 everyone $ x_n $ are irrational, but that's an open problem. I have developed a methodology to achieve some partial results in this direction.

If a reasonable one $ x_n> 1 $ then exists, 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 part $ b_n / a_n $ and $ f_n $ is its fraction. We get from the expansion of the Riemann zeta function in the Stieltjes series

$$

f_2 approx. 0.37241 le f_n le f_ {n + 1}

<1 -? about 0.422785

$$

Where $ gamma $ is the Euler-Mascheroni constant. $ f_n $ Strictly increasing it follows that $ zeta (x) $ decreases sharply in our region of interest $ x ge x_2 $,

So 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 $ so that $ gcd (r_n, a_n) = 1 $ and $ 0.37241 < dfrac {r_n} {a_n} <0.422789 $,

Now for everyone $ a_n ge 1 $ we can eliminate the possible values of $ r_n $ through computer verification to increase the lower limit $ a_n $ As shown below.

**Example: Prove $ a_n ge $ 6 to the $ n ge 2 $,**

To the $ 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 $ be eliminated. To the $ a_n = 5 $ there is a 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 testing. We watch that

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

Since $ dfrac {2} {5} $ lies between two successive values of $ f_3 $ and $ f_4 $ is getting bigger, we'll never have a situation where $ f_n = dfrac {2} {5} $ for each $ n $, Therefore $ r_n = 2 $ is excluded for $ a_n = 5 $, The whole numbers $ 3 $ and $ 4 $ are **Witnesses for the elimination** of $ dfrac {2} {5} $, With this we have exhausted all possibilities $ a_n = 5 $ and so we can conclude if $ zeta (x_n) in N $ then $ a_n ge $ 6, We can also eliminate them $ a_n = 6.7, ldots $ and so on.

**Note**: How $ a_n $ increases the number of possible $ r_n $ increases. Everyone $ r_n $ must be removed individually with a witness to completely eliminate $ a_n $, So after a point $ a_n $ will have several witnesses.

**definition**: A witness for a particular integer pair $ (a, r), a> r $ is an integer $ w_ {a, r} $ so 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 this $ zeta Big (1 + dfrac {a} {x + r} Big) $ is not an integer for an integer $ x> a> r $, Through actual calculation, I found the witness for every allowed pair $ (a, r), a le 10 ^ 4 $ what that implies

If $ x_n> 1 $ is rational and $ zeta (x_n) in N $ then the difference between

the numerator and denominator of $ x_n $ is bigger than $ 10 ^ 4 $,

**Size of the witness**

I noticed that most integers have small witnesses, but some of them have big witnesses. I looked at the list of integers $ a $ whose greatest witness for some $ r $ is greater than all witnesses of all integers less than $ a $, Let us call them the maximum witnesses. The first values of the maximum witnesses are given below:

```
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-consecutive denominators in the further group expansion of $ gamma $ but generally each of these values of $ a $ are the denominators in ascending order for the **best rational approximation $ gamma $,** In other words, the closer it is $ dfrac {r} {a} $ to $ 1- gamma $, the bigger his testimony $ w_ {a, r} $ that eliminates mathematically $ a $ more difficult.

I finally come to my question.

**question**: The problem is finding positive integer solutions $ zeta (p / q) = u $ synonymous with the decision about the rationality of $ gamma $,

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