Find every function $ f : \mathbb Q ^ + \to \mathbb Q ^ + $ such that
$$ f ( x + 1 ) = f ( x ) + 1 , \forall x \in \mathbb Q ^ + $$
and
$$ f \left( x ^ 2 \right) = f ^ 2 ( x ) , \forall x \in \mathbb Q ^ + \text . $$
Here, $ f ^ 2 ( x ) = f ( x ) ^ 2 = f ( x ) f ( x ) $.
All I have managed doing is showing that $ f ( x ) = x ^ c $ with this procedure:
In the equation $f(x^2)=f^2(x)$, we set $ x = e ^ y $. Therefore,
$$ f \left( e ^ { 2 y } \right) = f ^ 2 ( e ^ y ) \implies \ln \Big( f \left( e ^ { 2 y } \right) \Big) = 2 \ln \big( f ( e ^ y ) \big) \text . \tag 1 \label 1 $$
Now let $ g ( y ) = \ln \big( f ( e ^ y ) \big) $. So from \eqref{1} we have
$ g ( 2 y ) = 2 g ( y ) $.
This is a special situation of the functional equation of Cauchy which gives us
$$ g ( y ) = c y \implies \ln \big( f ( e ^ y ) \big) = c y \implies f ( e ^ y ) = e ^ { c y } = ( e ^ y ) ^ c \implies f ( x ) = x ^ c \text . $$
Any idea on how to continue?
Best Answer
Notation $f^2(x)$ to mean $f(x)f(x)$ is a bit confusing, at least in the context of functional equations usually $f^2(x)=f(f(x))$ and $f(x)^2=f(x)f(x)$. Also I am not sure how to continue your progress, since at the very start you put $x=e^y$, while $e$ is irrational number... Anyway, this has been solved for $f:\mathbb{Q}^+\to \mathbb{Q}^+$ many times outside this site, see search results. Following is the typical approach.
Using the $f(x+1)=f(x)+1$ inductively, we see $$f(x+n)=f(x)+n\tag{*}$$ for $x \in \mathbb{Q}^{+},n \in \mathbb{Z}^{+}$.
Now let $r \in \mathbb{Q}^{+}$, $n \in \mathbb{Z}^{+}$, we could evaluate $f((r+n)^2)$ in two ways, using the identity $f(x^2)=f(x)^2$ (denote $(**)$) and also expanding the expression $(r+n)^2$ and then use $(*)$. However, $2rn+n^2$ is not a natural number in general, we need to choose $n$ divisible by the denominator of $r$. So let $r=p/q$ and $n=q$ with $p,q \in \mathbb{Z}^{+}$. Then
$$f\left(\left(\frac{p}{q}+q\right)^2\right)\stackrel{(**)}{=}f\left(\frac{p}{q}+q\right)^2\stackrel{(*)}{=}\left(f\left(\frac{p}{q}\right)+q\right)^2\\=f\left(\frac{p}{q}\right)^2+2f\left(\frac{p}{q}\right)q+q^2,\tag{1}$$ but also $$f\left(\left(\frac{p}{q}+q\right)^2\right)=f\left(\left(\frac{p}{q}\right)^2+2p+q^2\right)\stackrel{(*)}{=}f\left(\left(\frac{p}{q}\right)^2\right)+2p+q^2\\\stackrel{(**)}{=}f\left(\frac{p}{q}\right)^2+2p+q^2.\tag{2}$$ Equating $(1)$ and $(2)$, we have $2f(\frac{p}{q})q=2p$, and so $f(\frac{p}{q})=\frac{p}{q}$. Since $p,q$ were arbitrary with $p/q \in \mathbb{Q}^{+}$, $f(x)=x$ is the only solution.