Find all functions f:R→R such that f(x3)+f(y3)=(x+y)f(x2)+f(y2)−f(xy) for all x∈R.
Problem
Source: Albanian National Math Olympiad 2012
Tags: function, blogs, algebra unsolved, algebra
01.04.2012 17:03
mhet49 wrote: Find all functions f:R→R such that f(x3) + f(y3) = (x+y)f(x2)+ f(y2) - f(xy) for all x ∈ R. Take x=y=0 than f(0)=0 If x=0 than f(y3)=f(y2) ..(a) If y=0 than f(x3)=xf(x2) or f(y3)=yf(y2) than for all z=y2≥0 we have f(z)=0 and from ..(a) f(x)=0 ,for all x∈R
01.04.2012 23:03
Bravo Arber.It is easy .I got the same result but i used a bit a different method
04.04.2012 21:20
mhet49, can you please confirm you meant what you posted? You're actual post was:
Which I edited to:
I have a feeling the problem should have been: mhet49 wrote: Find all functions f:R→R such that f(x3)+f(y3)=(x+y)(f(x2)+f(y2)−f(xy)) for all x∈R.
04.04.2012 21:27
Anyway, here is my solution to problem that I think is intended. Let P(x,−x) be the assertion that f(x3)+f(y3)=(x+y)(f(x2)+f(y2)−f(xy)). Then P(3√x,3√−x)⟹f(x)=−f(−x). Also, P(0,0)⟹f(0)=0. Therefore, P(x,0)⟹f(x3)=xf(x2). Expanding f(x3)+f(y3)=(x+y)(f(x2)+f(y2)−f(xy)) and using f(x3)=xf(x2), we see that (x+y)f(xy)=xf(y2)+yf(x2) which we will call the assertion Q(x,y). Then, Q(x,1)⟹(x+1)f(x)=xf(1)+f(x2) and Q(x,−1)⟹(x−1)(−f(x))=xf(1)−f(x2) (since f(x)=−f(−x)). Adding these two equations implies that 2f(x)=2xf(1) which gives the solution f(x)=cx for any real constant c.
04.04.2012 21:35
WakeUp wrote: I have a feeling the problem should have been: mhet49 wrote: Find all functions f:R→R such that f(x3)+f(y3)=(x+y)(f(x2)+f(y2)−f(xy)) for all x∈R. In this case, we can solve it as follows: Let x=y=0 to see f(0)=0. Then P(x,0)⟹f(x3)=xf(x2) and P(x,−x)⟹f(n)=−f(−n) for all real n. Then P(x,−x)⟹f(x3)−f(y3)=(x−y)(f(x2)+f(y2)+f(xy)). Adding this to the original and using f(x3)=xf(x2) gives yf(xy)=xf(y2), and plugging y=1 into this gives f(x)=xf(1)=ax for a constant a. We see that this always works, so f(x)=xf(1) is the solution. Edit: beaten, but this solution is a little different. Edit2: Wow, this matches up exactly with the solution in the below link..
04.04.2012 21:44
As I was reading your solution, I realised I found it familiar.. That's when I realised this problem is from the British Maths Olympiad Round 2 2009 Posted at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=254396, for example.
05.04.2012 21:59
i have use the same way as u.But in the end we can find that f(1)=0.
07.04.2012 14:32
WakeUp wrote: mhet49, can you please confirm you meant what you posted? You're actual post was:
Which I edited to:
I have a feeling the problem should have been: mhet49 wrote: Find all functions f:R→R such that f(x3)+f(y3)=(x+y)(f(x2)+f(y2)−f(xy)) for all x∈R. Actually no, it is just like what I have written it.
07.04.2012 14:53
mhet49 wrote: Find all functions f:R→R such that f(x3)+f(y3)=(x+y)f(x2)+f(y2)−f(xy) for all x∈R. My solution was as follows: f(x^{3})+ f(y^{3})= (x+y)f(x^{2})+ f(y^{2})- f(xy) the eqn for x=0 and y=o gives us f(0)=0 which we'll use for getting many valuable equations: first plug x=0, then we'll have f(y^{3})= f(y^{2}) whilst by plugging y=0 we get f(x^{3})= x f(x^{2}) so it means that f(x^{2})=0; also in the eqn for y=1 we have f(x)=f(x^{2}) which means f(x)=0 is the solution.
07.04.2012 15:23
mhet49 wrote: Actually no, it is just like what I have written it. Then I bet the proposers messed up the bracketing - can you ask them?
17.04.2012 22:18
mavropnevma wrote: mhet49 wrote: Actually no, it is just like what I have written it. Then I bet the proposers messed up the bracketing - can you ask them? They had!! This is what we had in the competition and what they corrected was with bracket like u said.... Unfortunate us
15.05.2012 19:43
Setting x=y= in the equation we get f(0)=0. Then setting y=0 we get f(x3)=xf(x2); in particular, putting x↦−x and adding we get f(−x3)=−f(x3) so f is odd. Finally, substituting y↦−y in the original equation and then adding it to the original equation we have yf(xy)=xf(y2), which for y=1 gives f(x)=f(1)x. Conversely, it is easily verified that, for any c∈R, f(x)=cx satisfies the conditions.
16.05.2012 01:08
Cassius, that is the exact same solution as applepi2000's, there's really no need to post it again.. you've also done this in a couple of other threads
16.05.2012 15:18
Dear WakeUp, I first read the problems on the resources page and solve them, then post my solutions on the threads and read others' after, so it may happen that a similar one was already posted. Moreover, I do it primarily also as an exercise to learn to write solutions since I have much need for it, so I will be glad if you will point out any mistake or omissions in mine. Having said that, I will avoid to post solutions to problems already solved, and please warn me if I do that error again.
17.05.2012 21:12
Well it seems fair enough, I don't mean to nag or be annoying, feel free to do so if you wish. But, for example, in this case, each step of your proof is identical to applepi2000's, and there is not much point spamming a thread with the same proof (it reminds me of the very first IMO problem, where nearly 10 equivalent proofs have been posted... thankfully that thread has been locked now though. In fact, it might have been me that locked it, I can't remember ). I suppose it is good practice for writing up solutions... but if there is already an identical proof posted, maybe try to avoid it! An alternative could be to post it on your blog or something..
18.05.2012 18:46
So f(x3)+f(y3)=(x+y)(f(x2)+f(y2)−f(xy)). Plugging in x=0, f(0)+f(y3)=yf(0)+yf(y2)−yf(0)=yf(y2). Now, y=0 yields 2f(0)=0, so f(0)=0. Thus, f(y3)=yf(y2), and f(−y3)=−yf(y2). Therefore, f(x) is odd (we could have found this with y=−x, but whatever). Thus, xf(x2)+yf(y2)=xf(x2)+yf(y2)+yf(x2)+xf(y2)−xf(xy)−yf(xy). Then, yf(x2)+xf(y2)=(x+y)f(xy). Letting y=1, f(x2)+xf(1)=(x+1)f(x). Also, y=−1 yields −f(x2)+xf(1)=(x−1)f(−x). Adding the two, f(x2)+f(−x2)+xf(1)+xf(1)=2xf(1)=xf(x)+xf(−x)+f(x)−f(−x)=2f(x). Then 2f(x)=2xf(1), so f(x)=xf(1). Thus, f(x)=ax. We have yf(x2)+xf(y2)=(x+y)f(xy), so yf(x2)+xf(y2)=ax2y+axy2=axy(x+y)=(x+y)f(xy). Thus, the solutions are all of the form f(x)=ax, and anything of that form is a solution. Q.E.D.
23.01.2017 20:30
My solution (WITHOUT BRACKETS)..... Putting x = y = 0, we get f(0)=0. Putting y = 0, we get f(x3)=xf(x2). Putting x =0, we get f(y3)=f(y2). Solving, we get f(y2)=0, provided y ≠ 1. Putting y = 1, we get f(x2)=f(x). This implies that f is even. So, f(1)=f(−1)=0. Hence, f(x)=0 for all x∈R.
12.04.2017 18:24
Plugging in x=y=0, we have that 2f(0)=f(0)−f(0)⟹f(0)=0. Plugging in x=0, we have f(0)+f(y3)=(y)f(0)+f(y2)−f(0)⟹f(y3)=f(y2). Plugging in y=0, we see that f(x3)+f(0)=(x)f(x2)+f(0)−f(0)⟹f(x3)=xf(x2). Comparing the last two statements, we see have that xf(x2)=f(x3)=f(x2)⟹xf(x2)=f(x2), implying that x is always zero, or that f(x) is always zero, of which the former is obviously false, hence f(x)=0. ◻
14.04.2017 21:02
Let P(x,y) be the assertion that f(x3)+f(y3)=(x+y)(f(x2)+f(y2)−f(xy)). P(x,−x),f(x)+f(−x)=0 so f(0)=0 P(x,0),f(x3)=xf(x2) and P(x,y)⟹(x+y)f(xy)=yf(x2)+xf(y2) (x+y)f(xy)=yf(x2)+xf(y2) above set y=1,(x+1)f(x)=f(x2)+xf(1) and set y=−1,(x−1)f(x)=−f(x2)+xf(1)⟹f(x)=xf(1) for all x∈R
14.04.2017 22:06
MonsterS wrote: Let P(x,y) be the assertion that f(x3)+f(y3)=(x+y)(f(x2)+f(y2)−f(xy)). P(x,−x),f(x)+f(−x)=0 so f(0)=0 P(x,0),f(x3)=xf(x2) and P(x,y)⟹(x+y)f(xy)=yf(x2)+xf(y2) (x+y)f(xy)=yf(x^2)+xf(y^2) above set y=1,(x+1)f(x)=f(x2)+xf(1) and set y=−1,(x−1)f(x)=−f(x2)+xf(1)⟹f(x)=xf(1) for all x∈R very nice solution
17.03.2018 12:31
x=0=y⟹f(0)=0 x=y⟹f(x3)=xf(x2) So coming back to the first equation xf(x2)+yf(y2)=(x+y)f(x2)+f(y2)−f(xy) Put above y=1 and get f(x2)=f(x) Now put to the given equation y=0 and get f(x3)=f(x2) So for all x∈R we have f(x2)=xf(x2)⟺0=(x−1)f(x2)⟺0=(x−1)f(x) That means x∈R∧x≠1⟹f(x)=0 Using f(x3)=xf(x2) put x=−1 and get 0=f(−1)=−f(1)⟹f(1)=0 So the solution is ∀x∈R f(x)=0. It's easy to check that it satisfies given conditions.
05.01.2019 07:30
Ez. Note that x=y=0 gives f(0)=0. Take y=0 to get f(x3)=xf(x2). Additionally, x=0 gives f(y3)=f(y2). Recalling f(y3)=yf(y2), we get f(y2)=0 for all y≠1. Thus, f(y3)=yf(y2)=0, which means f=0 for all x≠1. Now take P(1,2) to get f(1)=0 as well. Thus f=0
05.11.2020 21:13
By applying symmetry we see that x+y=1 and then we get the only solution as f(x) =0
14.08.2021 14:02
P(0,0):f(0)=0 P(x,0):f(x3)=xf(x2) P(0,y):f(y3)=f(y2) so f(x3)=xf(x3) and since any real number can be represented as a cube of another real, we must have f(x)=0 for all real x except 1, but since f is even f(1)=0 follows from 0=f(−1)=f(1) and so f(x)=x for all real x.
14.08.2021 20:15
P(0,0)⇒f(0)=0 P(x,0)⇒f(x3)=xf(x2) P(0,x)⇒f(x3)=f(x2)⇒(1−x)f(x2)=0, so f(x)=0∀x∈[0,1)∪(1,∞). Also, note that f is even, so it suffices to check f(1)=0. This holds because of P(1,−1).
17.08.2022 16:25
Let P(x,y) denote the assertion f(x3)+f(y3)=(x+y)(f(x2)+f(y2)−f(xy)). P(x,x)⟹f(x3)=xf(x2) P(x,−x)⟹−f(x)=f(−x) Combining P(x,1) and P(x,−1) ⟹f(x)=cx which works.
26.03.2023 12:34
Can someone please confirm what the answer is?? Half of solutions get f(x)=0 and half of them f(x)=kx and I am very very confused
26.03.2023 14:14
postingunsolvedquestions wrote: Can someone please confirm what the answer is?? Half of solutions get f(x)=0 and half of them f(x)=kx and I am very very confused Maybe because half of posts solve original problem f(x3)+f(y3)=(x+y)f(x2)+f(y2)−f(xy) While other half solved modified problem f(x3)+f(y3)=(x+y)(f(x2)+f(y2)−f(xy))
15.05.2023 17:25
mhet49 wrote: Find all functions f:R→R such that f(x3)+f(y3)=(x+y)f(x2)+f(y2)−f(xy)for all x∈R. x=y=0 we have f(0)=0 x=0 we have f(x3)=f(x2) y=0 we have f(x3)=xf(x2) ⇒f(x)=0,∀x≥0 Replace: −f(xy)=0⇒f(x)=0, ∀x∈R
28.10.2023 12:12
f(x^3)+f(y^3) = (x+y)f(x^2)+f(y^2)-f(xy) P(x,x) -> 2f(x^3) = 2xf(x^2) -> f(x^3) = xf(x^2) ... (i) P(y,y) -> 2f(y^3) = 2yf(y^2) -> f(y^3) = yf(y^2) ... (ii) P(0,y) -> f(y^3) = f(y^2) -> P(x) = f(x^3) = f(x^2) ... (iii) Comparing (i), (ii) and (iii), f(x^2) = f(x^3) = xf(x^2) -> f(x^2)-xf(x^2) = 0 -> f(x^2)(1-x) = 0 Either x = 1 or f(x) = 0 Since, there x is an independent constant, f(x) is 0.
30.10.2023 06:17
Might be the same as many solution before P(0,0)⟹f(0)=0 P(x,0)⟹f(x3)=xf(x2) P(0,x)⟹xf(x2)=f(x3)=f(x2) so f(x2)=0⟹f(x3)=0⟹f(x)=0
20.02.2024 11:47
Answer : f(x) = 0. Solution : Let P(x,y) denote the given assertion. P(0,0)⇒f(0)=0. P(x,0)⇒f(x3)=xf(x2).P(x,1)⇒f(x2)=f(x). x↔y,y→0⇒xf(x)=f(x)⇒f(x)=0.
22.02.2024 16:01
mhet49 wrote: Find all functions f:R→R such that f(x3)+f(y3)=(x+y)f(x2)+f(y2)−f(xy)for all x∈R. OK, so assuming this is the problem my solution is as follows. First, P(0,0) gives us that f(0)=0. Then, P(x,0)⟹f(x3)=xf(x2)P(0,x)⟹f(x3)=f(x2)for all x∈R. Combining these two relations we have that xf(x2)=f(x3)=f(x2)for all x∈R. Thus, f(x2)=0 for all x≠1. But this means that f(x2)=0 for all x (we can consider x to be -1 to deal with the edge case when x=1). Thus, f(t)=0 for all t≥0. Then, for all x≥0, P(−x,−x)⟹2f(−x3)=−2f(x2)+f(x2)−f(x2)=−2f(x2)which implies that f(−x3)=−f(x2) for all x≥0. Thus, f(t)=0 for all t≤0 as well. In conclusion the only working solution is f≡0 and we are done.
02.04.2024 04:15
Let P(x,y):f(x3)+f(y3)=(x+y)f(x2)+f(y2)−f(xy). Now, we take P(0,0):2f(0)=f(0)−f(0)=0, we obtain f(0)=0 P(0,y):f(y3)=f(y2) In this case we can obtain taking y=−1 f(−1)=f(1)…(1) P(x,0):f(x3)=xf(x2), also taking y=−1, we obtain f(−1)=−f(1), from (1),−f(1)=f(1)⇒f(1)=0 Now, f(x2)=f(x3)=xf(x2)So 0=(x−1)f(x2)∀x∈R, if I take x≠1, so f(α)=0∀1≠α≥0 but f(1) is also 0, so f(α)=0 for all α≥0 remembering f(x3)=f(x2)=0 because x2≥0 for all x real number. x3 goes through all the reals Therefore f≡0. As desired.
30.09.2024 10:33
Corrected version wrote: Find all functions f:R→R such that f(x3)+f(y3)=(x+y)(f(x2)+f(y2)−f(xy))for all x∈R. Sketch: P(x,x) for f(x3)=xf(x2), P(x,−y)+P(x,y) works.
09.12.2024 22:26
P(x,x) gives ===> f(x3)=x∗f(x2) (1) Then from P(0,0) ===> f(0)=0 (2) P(0,x) ===> f(x3)=f(x2) (3) (3) and (1) finishes the problem f(x3)=f(x2)=x∗f(x2) ofc x is not a constant , so f(x)=0!