Leartia wrote: Determine all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for every $x,y \in \mathbb{R}$ $$f(x^{4}-y^{4})+4f(xy)^{2}=f(x^{4}+y^{4})$$ Let $P(x,y)$ be the assertion $f(x^4-y^4)+4f(xy)^2=f(x^4+y^4)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(0,x)$ $\implies$ $f(x^4)=f(-x^4)$ and so $f(x)$ is even $P(x,x)$ $\implies$ $f(2x^4)=4f(x^2)^2$ and so $f(x)\ge 0$ $\forall x$ Let then $h(x)=\sqrt{f(x)}$ nonnegative even function $P(x,x)$ becomes $h(2x^4)=2f(x^2)$ and so $f(x)=\frac 12h(2x^2)$ $\forall x\ge 0$, still true when $x<0$ $P(x,y)$ may be written now as : $f(x^4-y^4)+f(2x^2y^2)=f(x^4+y^4)$ (using the preiviously got $f(2x^4)=4f(x^2)^2$) $h(2(x^4-y^4)^2)+h(8x^4y^4)=h(2(x^4+y^4)^2)$ $h((x-y)^2)+h(4xy)=h((x+y)^2)$ $\forall x,y\ge 0$ And so $h(u)+h(v)=h(u+v)$ $\forall u,v\ge 0$ This implies $h(x)=cx$ $\forall x\ge 0$ (additive + nonnegative) where $c=h(1)\ge 0$ And so $f(x)=\frac 12h(2x^2)=cx^2$ $\forall x$ Plugging this back in original equation, we get $c\in\{0,1\}$ and so : $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ and $\boxed{\text{S2 : }f(x)=x^2\quad\forall x}$

This problem was proposed by me.

dangerousliri wrote: This problem was proposed by me. How do you create FE problems that are good quality/suitable for Olympiads?

Cycle wrote: dangerousliri wrote: This problem was proposed by me. How do you create FE problems that are good quality/suitable for Olympiads? Well first I want to create a problem that a function which I want it to be the solution and I try to do it so that it does look good, then I try to solve it, if in the way of solving I find another solution it doesn't affect the problem. When I solve a problem I try again if I can solve shorter way. If I don't find a shorter solution and If I think the solution is good for test then I save that problem. For example the problem 5 in IMO 2015 that is what happend.

pco wrote: Leartia wrote: Determine all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for every $x,y \in \mathbb{R}$ $$f(x^{4}-y^{4})+4f(xy)^{2}=f(x^{4}+y^{4})$$ Let $P(x,y)$ be the assertion $f(x^4-y^4)+4f(xy)^2=f(x^4+y^4)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(0,x)$ $\implies$ $f(x^4)=f(-x^4)$ and so $f(x)$ is even $P(x,x)$ $\implies$ $f(2x^4)=4f(x^2)^2$ and so $f(x)\ge 0$ $\forall x$ Let then $h(x)=\sqrt{f(x)}$ nonnegative even function $P(x,x)$ becomes $h(2x^4)=2f(x^2)$ and so $f(x)=\frac 12h(2x^2)$ $\forall x\ge 0$, still true when $x<0$ $P(x,y)$ may be written now as : $f(x^4-y^4)+f(2x^2y^2)=f(x^4+y^4)$ (using the preiviously got $f(2x^4)=4f(x^2)^2$) $h(2(x^4-y^4)^2)+h(8x^4y^4)=h(2(x^4+y^4)^2)$ $h((x-y)^2)+h(4xy)=h((x+y)^2)$ $\forall x,y\ge 0$ And so $h(u)+h(v)=h(u+v)$ $\forall u,v\ge 0$ This implies $h(x)=cx$ $\forall x\ge 0$ (additive + nonnegative) where $c=h(1)\ge 0$ And so $f(x)=\frac 12h(2x^2)=cx^2$ $\forall x$ Plugging this back in original equation, we get $c\in\{0,1\}$ and so : $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ and $\boxed{\text{S2 : }f(x)=x^2\quad\forall x}$ Why from this $h(x)=\sqrt{f(x)}$ we get $h(2x^4)=2f(x^2)$ ?

Because of $h(x)=\sqrt{f(x)}$ when we have $f(2x^4)=4f(x^2)^2$ and take square roots both sides we have $\sqrt{f(2x^4)}=h(2x^4)=2f(x^2)$ I think this is the reason why.

Yes, I got it. Thanks a lot.

pco wrote: Leartia wrote: Determine all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for every $x,y \in \mathbb{R}$ $$f(x^{4}-y^{4})+4f(xy)^{2}=f(x^{4}+y^{4})$$ Let $P(x,y)$ be the assertion $f(x^4-y^4)+4f(xy)^2=f(x^4+y^4)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(0,x)$ $\implies$ $f(x^4)=f(-x^4)$ and so $f(x)$ is even $P(x,x)$ $\implies$ $f(2x^4)=4f(x^2)^2$ and so $f(x)\ge 0$ $\forall x$ Let then $h(x)=\sqrt{f(x)}$ nonnegative even function $P(x,x)$ becomes $h(2x^4)=2f(x^2)$ and so $f(x)=\frac 12h(2x^2)$ $\forall x\ge 0$, still true when $x<0$ $P(x,y)$ may be written now as : $f(x^4-y^4)+f(2x^2y^2)=f(x^4+y^4)$ (using the preiviously got $f(2x^4)=4f(x^2)^2$) $h(2(x^4-y^4)^2)+h(8x^4y^4)=h(2(x^4+y^4)^2)$ $h((x-y)^2)+h(4xy)=h((x+y)^2)$ $\forall x,y\ge 0$ And so $h(u)+h(v)=h(u+v)$ $\forall u,v\ge 0$ This implies $h(x)=cx$ $\forall x\ge 0$ (additive + nonnegative) where $c=h(1)\ge 0$ And so $f(x)=\frac 12h(2x^2)=cx^2$ $\forall x$ Plugging this back in original equation, we get $c\in\{0,1\}$ and so : $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ and $\boxed{\text{S2 : }f(x)=x^2\quad\forall x}$ The Quote: And so $h(u)+h(v)=h(u+v)$ $\forall u,v\ge 0$ This implies $h(x)=cx$ $\forall x\ge 0$ It is true only for continuous functions!

I am not sure about this but in a link about functional equation from evan chen there says that a function f is linear if any of these is true: 1)f is countinous in any interval 2)f is bounded (either above or below) in any nontrivial interval. 3)There exists (a, b) and ε > 0 such that (x − a)^2 + (f(x) − b)^2 > ε for every x So I think pco solution is correct because h(x) is bounded.

@2Above $h(x)$ being non-negative for all $x\in\mathbb{R}$ is also a sufficient condition for concluding that $h(x)$ is linear on non-negative domain.

Could we also do function switch like this: $f(x)=g(x^2)$? Someone, please verify I am not sure. And then since$$f(x^{4}-y^{4})+f(2x^{2}y^{2})=f(x^{4}+y^{4})$$this goes to$$g(x^{8}-2x^{4}y^{4}+y^{8})+g(4x^{4}y^{4})=g(x^{8}+2x^{4}y^{4}+y^{8})\implies g(a)+g(b)=g(a+b)\forall a,b\geq 0$$and thus since $g$ is bounded below by $0$, we conclude $g(x)=cx$ and thus $f(x)=f(x^2)=cx^2$ for all positive $x$, and since $f$ is even, our functions are in form $f(x)=cx^2$ and plugging back in, we get that $f(x)=0$ or $f(x)=x^2$.

rafaello wrote: Could we also do function switch like this: $f(x)=g(x^2)$? Someone, please verify I am not sure. And then since$$f(x^{4}-y^{4})+f(2x^{2}y^{2})=f(x^{4}+y^{4})$$this goes to$$g(x^{8}-2x^{4}y^{4}+y^{8})+g(4x^{4}y^{4})=g(x^{8}+2x^{4}y^{4}+y^{8})\implies g(a)+g(b)=g(a+b)\forall a,b\geq 0$$and thus since $g$ is bounded below by $0$, we conclude $g(x)=cx$ and thus $f(x)=f(x^2)=cx^2$ for all positive $x$, and since $f$ is even, our functions are in form $f(x)=cx^2$ and plugging back in, we get that $f(x)=0$ or $f(x)=x^2$. shouldn't u first prove that the function is even?(and then $f(x)=g(x^2)$)

iman007 wrote: rafaello wrote: Could we also do function switch like this: $f(x)=g(x^2)$? Someone, please verify I am not sure. And then since$$f(x^{4}-y^{4})+f(2x^{2}y^{2})=f(x^{4}+y^{4})$$this goes to$$g(x^{8}-2x^{4}y^{4}+y^{8})+g(4x^{4}y^{4})=g(x^{8}+2x^{4}y^{4}+y^{8})\implies g(a)+g(b)=g(a+b)\forall a,b\geq 0$$and thus since $g$ is bounded below by $0$, we conclude $g(x)=cx$ and thus $f(x)=f(x^2)=cx^2$ for all positive $x$, and since $f$ is even, our functions are in form $f(x)=cx^2$ and plugging back in, we get that $f(x)=0$ or $f(x)=x^2$. shouldn't u first prove that the function is even?(and then $f(x)=g(x^2)$) That is trivial without function swap. My main doubt is that can we take function so that $f(x)=g(x^2)$, is it valid, it seems to be valid, but I am not so sure.

As already said in post #17, you can not write $f(x)=g(x^2)$ for some $g(x)$ if you did not prove before that $f(x)$ is an even function.

Ok $P(0,0)+P(0,x)\implies f \text{ is even}$, thus this function swap is valid, thanks. So I guess, my solution is easier