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It seems to me that the functions $f(x) = \left(1 + \frac {1}{x}\right)^x$ and $g (x) = \left(1 + \frac {1}{x}\right)^{x+1}$, defined for every positive real $x$, assume simultaneously rational values if, and only if, $x$ is integer. I tested by assigning values to $x$, but I could not elaborate proof of this statement.

Suppose that f (x) and g (x) are rational. Let A, B, C and D be positive integers, with gcd (A, B) = gcd (C, D) = 1. Let f (x) = A / B and g (x) = C / D.

So:

C / D = (A / B) (1 + 1 / x) and, therefore, after the calculations:

x = (BC-AD) / AD

Well, now I would like to conclude that x is integer, but I do not know if it is possible.

I need help.

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    That can't be true. $f(1)=2,f(2)=2.15166$ and the function is continuous so it hits every rational number between. – 2017-01-09
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    @lulu Oh. Well darn, I'm stupid for trying... – 2017-01-09
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    This becomes much more interesting if you restrict the domain of the function to rational numbers! – 2017-01-09
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    @SimpleArt Well, maybe $f(x)\in \mathbb Q$ for $x\notin \mathbb Z$ implies that $x$ is irrational? – 2017-01-09
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    @lulu Yes, that was just about where I was getting in the not-yet-existent answer I am trying to write up :-D – 2017-01-09
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    @lulu Stronger, I think it implies $x$ is transcendental by that one theorem about algebraic numbers raised to each other. – 2017-01-09
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    https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem – 2017-01-09
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    @SimpleArt I don't immediately see how that applies. Theres plenty of cases where $a,b,a^b$ are all rational. – 2017-01-09
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    f(1) = 2, f(2) = 9/4 – 2017-01-09
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    @lulu I meant that when $x$ is irrational, it must be transcendental, right? If $x$ were non-rational algebraic and $f(x)$ were rational, it would contradict the theorem. – 2017-01-09
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    @PauloArgolo You are correct, I was giving $f(1.5)$ not $f(2)$. But it doesn't really matter what the values are...the point is that it must hit all the rationals between the two points. For example, we see that there is some $x_0$ with $1$f(x_0)=2.1$ and clearly $x_0\notin \mathbb Z$. – 2017-01-09
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    @SimpleArt Ah, that sounds right. – 2017-01-09
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    @PauloArgolo Wait...were you saying that $f(x)$ and $g(x)$ simultaneously rational should imply that $x\in \mathbb Z$? I thought you were asking two separate questions. of course if both of these values are rational then the quotient $\frac {g(x)}{f(x)}=(1+\frac 1x)$ is rational which at least implies that $x$ is rational. – 2017-01-09
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    @PauloArgolo With the simultaneous condition you are now asking the same question we've been asking in the comments. Namely, if $x,f(x),g(x)$ are simultaneously rational does that imply that $x$ is an integer? – 2017-01-09

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If $x\in\mathbb N$, then

$$f(x)=\frac{(x+1)^x}{x^x}$$

which is trivially rational.

Assume that $x$ is a non-integer rational number of the form $a/b$, where $a,b\in\mathbb N$ and it is in reduced form. Then, if we are to have $f(x)$ be rational,

$$f(a/b)=\left(\frac{a+b}a\right)^{a/b}=\frac cd$$

$$\implies\begin{cases}(a+b)^a=c^b\\a^a=d^b\end{cases}$$

From here, we can see that $d$ must factor into $a$, but this only occurs if $b$ has a common factor with $a$ so that the amount of factors on both sides is the same. Then we have contradiction, since $x$ must be in reduced form, which implies $a$ does not divide $b$.

By the Gelfond-Schneider theorem, if $x$ is algebraic and not rational, then $f(x)$ is transcendental, which is definitely not what we want.

Thus, the only solutions are if $x$ is a natural number or if $x$ is transcendental. An argument of continuity will show the existence of these transcendental cases.

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    I had to edit the question. – 2017-01-09
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    I kept writing this down, but never convinced myself. Where, for example, do you use the fact that $b\neq 1$? – 2017-01-09
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    @lulu That $x$ is non-integer rational. First line. – 2017-01-09
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    @PauloArgolo Well, that changes it drastically. If lulu wants to answer that one, go ahead :D – 2017-01-09
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    Oddly, the simultaneous question turns out to be equivalent to the one we posed each other. Note that $\frac {g(x)}{f(x)}=(1+\frac 1x) $ so if they are both rational then so is $x$. – 2017-01-09
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    I see where you assumed $x$ was non-integral but I don't see where you used it. If I take $x=2$, where does your argument break down? I mean...you have $d^b=a^a$. How does that imply that $b$ divides $a$? $2^8=4^4$ but $8$ does not divide $4$. – 2017-01-09
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    @lulu I did not cover the cases for $x\in\mathbb N$, as I considered them trivial. Think I should add? – 2017-01-09
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    The integer case is trivial...But I'm just not following your logic. Like I say, it is not true that $d^b=a^a\implies b\,|\,a$ as the example $2^8=4^4$ shows. – 2017-01-09
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    @lulu Oh, my bad. I meant that $a$ and $b$ had common factors :-) – 2017-01-09
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    Ok. if $p$ is a prime dividing $b$ then it is certainly true that what you have shows that $p\,|\,a$ . I'm convinced. Thanks! – 2017-01-09
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    @lulu And thanks for clearing things up! :D I'm actually horrible at this stuff IMO, its not my "tag" of expertise. – 2017-01-09
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    Alas, the argument still needs some work. I was wrong...it isn't true that $d^b=a^a$ means that $a,b$ have to have factors in common. Take $a=2^3$, so $a^a=2^{3a}=2^{24}$. Then we could have $d=2^8,b=3$. I keep making versions of the same error on this problem...I'll think about it later in the day. – 2017-01-09