Daily

Random

Practice set

Baltic Way 1998 · Problem 7

Algebra

Let R\mathbb{R} be the set of all real numbers. Find all functions f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} satisfying for all x,y∈Rx, y \in \mathbb{R} the equation

f(x)+f(y)=f(f(x)f(y)).f(x)+f(y)=f(f(x) f(y)) .
Change pool

When you’re ready

Review material becomes available with the next Daily.

Review

Topics

Functional equations

Solutions

Solution

Answer: f(x)≡0f(x) \equiv 0 is the only such function.

Choose an arbitrary real number x0x_{0} and denote f(x0)=cf\left(x_{0}\right)=c. Setting x=y=x0x=y=x_{0} in the equation we obtain f(c2)=2cf\left(c^{2}\right)=2 c. For x=y=c2x=y=c^{2} the equation now gives f(4c2)=4cf\left(4 c^{2}\right)=4 c. On the other hand, substituting x=x0x=x_{0} and y=4c2y=4 c^{2} we obtain f(4c2)=5cf\left(4 c^{2}\right)=5 c. Hence 4c=5c4 c=5 c, implying c=0c=0. As x0x_{0} was chosen arbitrarily, we have f(x)=0f(x)=0 for all real numbers xx.

Obviously, the function f(x)≡0f(x) \equiv 0 satisfies the equation. So it is the only solution.

Contest context

Results from Baltic Way 1998

11 teams

Mean score
2.8 / 5
Scores of 4 or 5
5 / 11
Estonia
1 / 5

Score distribution

01
14
21
30
40
55
All team scores
TeamScore
Latvia5 / 5
Estonia1 / 5
Poland5 / 5
Finland5 / 5
St. Petersburg1 / 5
Sweden5 / 5
Denmark1 / 5
Iceland5 / 5
Norway1 / 5
Germany2 / 5
Lithuania0 / 5