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Baltic Way 2017 · Problem 5

Algebra

Find all functions f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} such that

f(x2y)=f(xy)+yf(f(x)+y)f\left(x^{2} y\right)=f(x y)+y f(f(x)+y)

for all real numbers xx and yy.

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Topics

Functional equations · Sequences and recurrences

Solutions

Solution

Answer: f(y)=0f(y)=0.

By substituting x=0x=0 into the original equation we obtain f(0)=f(0)+yf(f(0)+y)f(0)=f(0)+y f(f(0)+y), which after simplifying yields yf(f(0)+y f(f(0)+ y)=0y)=0. This has to hold for every yy.

Now let's substitute y=−f(0)y=-f(0). We get that −(f(0))2=0-(f(0))^{2}=0, which gives us f(0)=0f(0)=0.

By plugging the last result into yf(f(0)+y)=0y f(f(0)+y)=0 we now get that yf(y)=0y f(y)=0.

Therefore if y≠0y \neq 0 then f(y)=0f(y)=0.

Altogether we have shown that f(y)=0f(y)=0 for every yy is the only possible solution, and it clearly is a solution.

Contest context

Results from Baltic Way 2017

11 teams

Mean score
4.7 / 5
Scores of 4 or 5
11 / 11
Estonia
4 / 5

Score distribution

00
10
20
30
43
58
All team scores
TeamScore
St. Petersburg5 / 5
Germany5 / 5
Poland5 / 5
Denmark4 / 5
Estonia4 / 5
Lithuania5 / 5
Sweden5 / 5
Norway5 / 5
Finland4 / 5
Iceland5 / 5
Latvia5 / 5