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Algebra Difficulty 6.3 National olympiad Find the answer

Determine all functions f:(0,)Rf:(0, \infty) \rightarrow \mathbb{R} satisfying
(x+1x)f(y)=f(xy)+f(yx) \left(x+\frac{1}{x}\right) f(y)=f(x y)+f\left(\frac{y}{x}\right)
for all x,y>0x, y>0. (South Korea)
Answer: f(x)=C1x+C2xf(x)=C_{1} x+\frac{C_{2}}{x} with arbitrary constants C1C_{1} and C2C_{2}.

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

Solution

Fix a real number a>1a>1, and take a new variable tt. For the values f(t),f(t2)f(t), f\left(t^{2}\right), f(at)f(a t) and f(a2t2)f\left(a^{2} t^{2}\right), the relation (1) provides a system of linear equations:
x=y=t:(t+1t)f(t)=f(t2)+f(1)x=ta,y=at:(ta+at)f(at)=f(t2)+f(a2)x=a2t,y=t:(a2t+1a2t)f(t)=f(a2t2)+f(1a2)x=y=at:(at+1at)f(at)=f(a2t2)+f(1) \begin{array}{llll} x=y=t: & \left(t+\frac{1}{t}\right) f(t) & =f\left(t^{2}\right)+f(1) \\ x=\frac{t}{a}, y=a t: & \left(\frac{t}{a}+\frac{a}{t}\right) f(a t) & =f\left(t^{2}\right)+f\left(a^{2}\right) \\ x=a^{2} t, y=t: & \left(a^{2} t+\frac{1}{a^{2} t}\right) f(t) & =f\left(a^{2} t^{2}\right)+f\left(\frac{1}{a^{2}}\right) \\ x=y=a t: & \left(a t+\frac{1}{a t}\right) f(a t) & =f\left(a^{2} t^{2}\right)+f(1) \end{array}
In order to eliminate f(t2)f\left(t^{2}\right), take the difference of (2a) and (2b); from (2c) and (2d) eliminate f(a2t2)f\left(a^{2} t^{2}\right); then by taking a linear combination, eliminate f(at)f(a t) as well:
(t+1t)f(t)(ta+at)f(at)=f(1)f(a2) and (a2t+1a2t)f(t)(at+1at)f(at)=f(1/a2)f(1), so ((at+1at)(t+1t)(ta+at)(a2t+1a2t))f(t)=(at+1at)(f(1)f(a2))(ta+at)(f(1/a2)f(1)). \begin{gathered} \left(t+\frac{1}{t}\right) f(t)-\left(\frac{t}{a}+\frac{a}{t}\right) f(a t)=f(1)-f\left(a^{2}\right) \text { and } \\ \left(a^{2} t+\frac{1}{a^{2} t}\right) f(t)-\left(a t+\frac{1}{a t}\right) f(a t)=f\left(1 / a^{2}\right)-f(1), \text { so } \\ \left(\left(a t+\frac{1}{a t}\right)\left(t+\frac{1}{t}\right)-\left(\frac{t}{a}+\frac{a}{t}\right)\left(a^{2} t+\frac{1}{a^{2} t}\right)\right) f(t) \\ =\left(a t+\frac{1}{a t}\right)\left(f(1)-f\left(a^{2}\right)\right)-\left(\frac{t}{a}+\frac{a}{t}\right)\left(f\left(1 / a^{2}\right)-f(1)\right) . \end{gathered}
Notice that on the left-hand side, the coefficient of f(t)f(t) is nonzero and does not depend on tt:
(at+1at)(t+1t)(ta+at)(a2t+1a2t)=a+1a(a3+1a3)<0. \left(a t+\frac{1}{a t}\right)\left(t+\frac{1}{t}\right)-\left(\frac{t}{a}+\frac{a}{t}\right)\left(a^{2} t+\frac{1}{a^{2} t}\right)=a+\frac{1}{a}-\left(a^{3}+\frac{1}{a^{3}}\right)<0 .
After dividing by this fixed number, we get
f(t)=C1t+C2t f(t)=C_{1} t+\frac{C_{2}}{t}
where the numbers C1C_{1} and C2C_{2} are expressed in terms of a,f(1),f(a2)a, f(1), f\left(a^{2}\right) and f(1/a2)f\left(1 / a^{2}\right), and they do not depend on tt. The functions of the form (3) satisfy the equation:
(x+1x)f(y)=(x+1x)(C1y+C2y)=(C1xy+C2xy)+(C1yx+C2xy)=f(xy)+f(yx). \left(x+\frac{1}{x}\right) f(y)=\left(x+\frac{1}{x}\right)\left(C_{1} y+\frac{C_{2}}{y}\right)=\left(C_{1} x y+\frac{C_{2}}{x y}\right)+\left(C_{1} \frac{y}{x}+C_{2} \frac{x}{y}\right)=f(x y)+f\left(\frac{y}{x}\right) .

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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.