Determine all functions f:R→R such that for all x,y∈R we have: xf(x+y)+yf(y−x)=f(x2+y2).Proposed by Nikola Velov
Problem
Source: Macedonian Mathematical Olympiad 2023 P1
Tags: algebra, functional equation, Macedonia, national olympiad
09.04.2023 22:41
I claim all such functions are form f(x)=cx, where c∈R is arbitrary. Denote the given assertion by P(x,y). Note that P(0,y) gives f(y2)=yf(y). In particular f(0)=0 and f((−y)2)=−yf(−y) so that f(−y)=−f(y) for any y. Now, P(x,y)⇒xf(x+y)+yf(y−x)=f(x2+y2)P(y,x)⇒yf(x+y)+xf(x−y)=f(x2+y2).As f(y−x)=−f(x−y), we get (x−y)f(x+y)=(x+y)f(x−y)for any x,y. Now we are ready for conclusion. Take any r1,r2≠0 and let x,y be such that x+y=r1 and x−y=r2. We then get immediately f(r1)/r1=f(r2)/r2 for any r1,r2≠0. So, there is a c such that f(x)=cx for any x≠0. Clearly this extends to all x as f(0)=0, which completes the solution.
10.04.2023 16:19
a trivial problem just need to change x into -x.
10.04.2023 16:45
Natsuki_kobayashi_486 wrote: a trivial problem just need to change x into -x. That's why it's the first of the five problems.
10.04.2023 17:13
10.06.2023 02:05
steppewolf wrote: Determine all functions f:R→R such that for all x,y∈R we have: xf(x+y)+yf(y−x)=f(x2+y2).Proposed by Nikola Velov xf(x+y)+yf(y−x)=f(x2+y2)...(α)In (α)x=y=0: ⇒f(0)=0In (α)x=0: ⇒yf(y)=f(y2)⇒f is oddIn (α)x=y,y=x: yf(x+y)+xf(x−y)=f(x2+y2)...(β)By (α) and (β): ⇒yf(x+y)+xf(x−y)=xf(x+y)+yf(y−x)⇒(x+y)f(x−y)=(x−y)f(x+y)x+y=a,x−y=b: ⇒af(b)=bf(a),∀a,b∈R...(θ)In (θ)a=1: ⇒f(b)=bf(1),∀b∈RReplacing in (α): f(x)=xc,c is constant and x∈R is the only solution