Auswahlklausur

Solution:

We begin with some properties of the base set $\mathbb{Q}^{+}$ that are important for the solution set. As is well known, the product $a b$ of two numbers $a, b \in \mathbb{Q}^{+}$ again lies in $\mathbb{Q}^{+}$, as does the inverse $\frac{1}{a}$ for every number $a \in \mathbb{Q}^{+}$ (which is not always satisfied for $\sqrt{a}$). Since the neutral element $1$ of the (commutative) multiplication also belongs to $\mathbb{Q}^{+}$, $\mathbb{Q}^{+}$ forms a commutative (= abelian) group with respect to multiplication. Every non-empty subset $U$ for which $a, b \in U$ always implies $\frac{a}{b} \in U$ is an (abelian) subgroup of $\mathbb{Q}^{+}$, i.e., it contains the number $1$, every product of two of its elements, and the reciprocal of each of its numbers. An example is $D=\left\{3^{t} \mid t \in \mathbb{Z}\right\}$, the set of all powers of three with integer exponents; trivial examples are $U=\{1\}$ and $U=\mathbb{Q}^{+}$.

Furthermore, for a subgroup $U$ of $\mathbb{Q}^{+}$ and an arbitrary number $a \in \mathbb{Q}^{+}$, the set $a U=\{a u \mid u \in U\}$ is called the coset of $\mathbb{Q}^{+} / U$ with representative $a$. The representative of a coset is generally not uniquely determined; likewise, the cosets themselves are generally not groups. An example with $a=2$ is the set $2 D=\left\{2 \cdot 3^{t} \mid t \in \mathbb{Z}\right\}$, which, for instance, also has the representative $6$.

After these preliminary remarks, we can now first describe the solution set:

Let $U$ be an arbitrary subgroup of $\mathbb{Q}^{+}$ and let $a_{1}, a_{2}, \ldots \in \mathbb{Q}^{+}$ be fixed representatives of the cosets of $\mathbb{Q}^{+} / U$. Furthermore, let $u_{1}, u_{2}, \ldots$ be numbers from $U$, such that exactly one $u_{i}$ is assigned to each $a_{i}$. This yields the following solutions $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$:
$$f(x)=\left\{\begin{array}{cl} r \cdot u_{i} & \text{ for } x=a_{i}^{2} \cdot r^{2} \text{ with } r \in U \\ u \text{ arbitrary } \in U & \text{ if } x \text{ is not the square of a number from } \mathbb{Q}^{+} \end{array}\right.$$
Here all solutions arise through the possible subgroups $U$ and the choice of the $u_{i}$, as well as through arbitrary choice of $u \in U$ (the $a_{i}$ being fixed for each $U$).

For example, the following results for the trivial subgroups of $\mathbb{Q}^{+}$:

a) $U=\{1\}$: Here only $r=u=1$ as well as $u_{i}=1$ is possible, so that $f(x)=1$ for all $x \in \mathbb{Q}^{+}$.

b) $U=\mathbb{Q}^{+}$: Here $f(x)=a \sqrt{x}$ with fixed $a$, when $x$ is a perfect square, and $f(x)$ arbitrary from $\mathbb{Q}^{+}$, when $x$ is not a perfect square.

The proof proceeds in two steps.

Step 1: We show that the given functions $f$ are well-defined and satisfy $(*)$.

Every $t=x^{2}$ with $x \in \mathbb{Q}^{+}$ can, by the definition of the cosets, be uniquely represented in the form $t=a_{i}^{2} \cdot r^{2}$ with $r \in U$, and every product $a_{i}^{2} \cdot r^{2}$ is a rational perfect square. Thus $f$ is well-defined. It further follows from the definition that the value set $W_{f}$ of $f$ lies in $U$. Therefore, for every $y \in \mathbb{Q}^{+}$ there exists a $u \in U$ with $f(y)=u$. Since, as noted above, for every $x \in \mathbb{Q}^{+}$ a unique representation $x^{2}=a_{i}^{2} \cdot r^{2}$ exists, it follows that
$$f\left(x^{2} \cdot f(y)^{2}\right)=f\left(a_{i}^{2} r^{2} u^{2}\right)=f\left(a_{i}^{2}(r u)^{2}\right)=(r u) \cdot u_{i}=\left(r u_{i}\right) \cdot u=f\left(a_{i}^{2} r^{2}\right) f(y)=f\left(x^{2}\right) f(y),$$
which shows that $(*)$ is satisfied.

Step 2: We show that no further solutions of $(*)$ can exist besides the ones given.

To this end, let $f$ be a solution function of $(*)$ with the solution set $W_{f} \subseteq \mathbb{Q}^{+}$.

We substitute $y=1$ and $x=\frac{1}{f(1)}$ into $(*)$:
$$f\left(x^{2} f(y)^{2}\right)=f\left(\frac{f(1)^{2}}{f(1)^{2}}\right)=f(1)=f\left(x^{2}\right) f(1).$$
Thus $1 \in W_{f}$.

Now let $z=\frac{1}{f(1)}$ be fixed (by the above, this means $f\left(z^{2}\right)=1$). Further let $w \in W_{f}$ (this means there exists a $u \in \mathbb{Q}^{+}$ with $f(u)=w$). We substitute $y=u$ and $x=\frac{z}{f(y)}$ into $(*)$:
$$f\left(x^{2} f(y)^{2}\right)=f\left(z^{2}\right)=1=f\left(x^{2}\right) \cdot w,$$
hence $f\left(x^{2}\right)=\frac{1}{w}$. Thus for every $w \in W_{f}$, $\frac{1}{w}$ also lies in $W_{f}$.

Now let $v, w \in W_{f}$ (this means there exist $t, u \in \mathbb{Q}^{+}$ with $f(t)=v$ and $f(u)=w$).
We substitute $x=z=\frac{1}{f(1)}$ and $y=u$ into $(*)$, denote $k=x \cdot f(y)$, and obtain $f\left(k^{2}\right)=1 \cdot f(u)=w$.
Then we substitute $x=k$ and $y=t$ anew into $(*)$:
$$f\left(x^{2} f(y)^{2}\right)=f\left(k^{2}\right) f(t)=w \cdot v.$$
Thus for $v, w \in W_{f}$, the product $v w$ always lies in $W_{f}$ as well.

This proves that $W_{f}$ is a subgroup of $\mathbb{Q}^{+}$.

Now let $a_{1}, a_{2}, \ldots$ be fixed chosen representatives of the cosets of $\mathbb{Q}^{+} / W_{f}$ (as in the description of the solution set) and let $r \in W_{f}$ be arbitrary with $f(u)=r$. We substitute $x=a_{i}$ (for arbitrary $i$) and $y=u$ into $(*)$:
$$f\left(x^{2} f(y)^{2}\right)=f\left(a_{i}^{2} \cdot r^{2}\right)=f\left(a_{i}^{2}\right) f(u)=u_{i} \cdot r$$
(where $f\left(a_{i}^{2}\right):=u_{i}$ is set). From this it is evident that every solution function $f$ must indeed have the form described above. $\square$