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Algebra Difficulty 6.8 National olympiad Find the answer

Example 39([31.4]) Let the set of all positive rational numbers be denoted as Q+Q^{+}. Construct a function f:Q+Q+f: Q^{+} \rightarrow Q^{+}, such that for any x,yQ+x, y \in Q^{+}, it satisfies yf(xf(y))=f(x)y f(x f(y))=f(x).

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Assume the function ff satisfies the given condition yf(xf(y))=f(x)y f(x f(y))=f(x). Let's first discuss the basic properties such a function ff should satisfy. Taking x=1x=1, from the condition we get
f(f(y))=f(1)/y.f(f(y))=f(1) / y.

If f(x)=f(y)f(x)=f(y), then from this and equation (1) we derive
f(1)/y=f(f(y))=f(f(x))=f(1)/x, i.e., x=y.f(1) / y=f(f(y))=f(f(x))=f(1) / x, \text{ i.e., } x=y.

Therefore,
f(x)=f(y)f(x)=f(y) if and only if x=yx=y.
Taking x=y=1x=y=1, from the condition we get f(f(1))=f(1)f(f(1))=f(1). From this and equations (2), (1), we have
f(1)=1,f(f(y))=1/y.f(1)=1, \quad f(f(y))=1 / y.

Substituting f(x)f(x) for xx, from the condition and equation (3) we get
f(f(x)f(y))=f(f(x))/y=1/(xy)=f(f(xy)).f(f(x) f(y))=f(f(x)) / y=1 /(x y)=f(f(x y)).

From this and equation (2), we derive
f(xy)=f(x)f(y),f(x y)=f(x) f(y),

i.e., f(x)f(x) is a "totally multiplicative function". Since f(x)f(x) is not equal to zero, it is easy to see (why) that the condition yf(xf(y))=f(x)y f(x f(y))=f(x) holds if and only if conditions (1) and (5) hold.

Now, the function ff is defined on Q+Q^{+}. By the fundamental theorem of arithmetic, xQ+x \in Q^{+} if and only if xx has the representation
x=p1a(1)pra(r),a(j)Z,1jr,x=p_{1}^{a(1)} \cdots p_{r}^{a(r)}, \quad a(j) \in \mathbb{Z}, 1 \leqslant j \leqslant r,

where p1,,pk,p_{1}, \cdots, p_{k}, \cdots are all prime numbers, arranged in increasing order. When xx is given by the above expression, from equation (5) and f(1)=1f(1)=1 we get
f(x)=f(p1)a(1)f(pr)a(r).f(x)=f\left(p_{1}\right)^{a(1)} \cdots f\left(p_{r}\right)^{a(r)}.

Therefore, it is sufficient to define the value of the function ff at prime numbers pp to meet the required conditions. From the above analysis, it is known that it is sufficient for prime numbers pp to satisfy condition (1), i.e.,
f(f(p))=1/p.f(f(p))=1 / p.

Since ff is a function from Q+Q+Q^{+} \rightarrow Q^{+}, we can assume
f(p)=q1a(1)qra(r),qj is a prime, a(j)Z,1jr.f(p)=q_{1}^{a(1)} \cdots q_{r}^{a(r)}, \quad q_{j} \text{ is a prime, } a(j) \in \mathbb{Z}, 1 \leqslant j \leqslant r.

From equation (5) we know
1/p=f(f(p))=f(q1)a(1)f(qr)a(r).1 / p=f(f(p))=f\left(q_{1}\right)^{a(1)} \cdots f\left(q_{r}\right)^{a(r)}.

Since f(1)=1f(1)=1 only when x=1x=1, the above equation suggests that f(p)f(p) should take a very simple value, such as qq or 1/q1 / q, where qq is a prime. However, these discussions do not truly provide information on how to define the value of f(p)f(p). Through experimentation and observation, for j1j \geqslant 1 we define
f(p2j1)=1/p2j,f(p2j)=p2j1f\left(p_{2 j-1}\right)=1 / p_{2 j}, \quad f\left(p_{2 j}\right)=p_{2 j-1}

to meet the requirements.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.