AoPS - Problem

Sardor

15.12.2014 12:39

It's not very hard! Here my longer solution: Let $ P(x,y) $ be the assertion of the function. From $ P(x,-y)=P(x,y) $, we get $ (f(y)-f(-y))(f(x+y)+f(x-y)+f(y)+f(-y) (*) $. We have $ P(0,0) \implies f(0)=0 $ or $ f(0)=\frac{1}{2} $. Case 1: $ f(0)=0 $ i) $ P(0,x) \implies f(2x^2)=2f(x)(f(x)+f(-x)) $ and $ P(x,0) \implies f(x^2)=f(x)^2 $, thus $ f(x)=f(-x) $ or $ f(x)=-f(-x) $. If $ f $ is odd, the $ f(2x^2)=0 $, so $ f(x)=0 $ for all $ x \in R_0 $ and since $ f $ is odd, we get $ f(x)=0 $, for any $ x $. Let $ f $ is even, then $ f(2x^2)=4f(x)^2=4f(x^2) $, so $ f(2x)=4f(x) $, for any $ x $ ( because $ f $ is even).Taking $ x=y $, in original equation, we get $ 3f(x)^2=2f(x)f(2x)=8f(x)^2 $ or $ f(x)=0 $, for any $ x $. Case 2: $ f(0)=\frac{1}{2} $. Then $ P(x,0) \implies f(x^2)=f(x)^2+f(x)-\frac{1}{4} $ , so $ f $ is even or $ f(x)+f(-x)+1=0 $ a) $ f(x)=f(-x) $. Then $ P(0,x) \implies f(2x^2)=4f(x)^2-\frac{1}{2} $.Hence $ (f(x+y)+f(y))(f(x-y)+f(y)=f(x^2)+f(2y^2)=f(x)^2+f(x)+4f(y)^2-\frac{3}{4} $.Then taking $ x=y $ in the least equation we have $ f(2x)=4f(x)-\frac{3}{2} $ (because $ f(x)=-\frac{1}{2} $ isn't solution).So we have $ 4f(x)^2-\frac{1}{2}=f(2x^2)=4f(x^2)-\frac{3}{2}=4f(x)^2+4f(x)-1-\frac{3}{2} $, so $ f(x)=\frac{1}{2} $ b) Let $ f(x)+f(-x)+1=0 $, then by $ (*) $, we get $ f(x+y)+f(x-y)=1 $ for $ y $ not equal to zero, and taking $ y=x $, we get $ f(x)=\frac{1}{2} $, for any $ x $, but $ 0=f(x)+f(-x)+1=2 $, a contradiction. Answer: 1) $ f(x)=0 $; 2) $ f(x)=\frac{1}{2} $;

pco

15.12.2014 13:14

Sardor wrote: It's not very hard! ... Answer: 1) $ f(x)=0 $; 2) $ f(x)=\frac{1}{2} $; And what about $f(x)=x^2$ for example ?

Mukhammadiev

15.12.2014 14:02

You are right, Sardor. My solution almost the same as your's. However, $ f(x)=x^2$ holds the aforementioned equation, the equation does not pay the way for this.

joyce_tan

16.12.2014 06:02

Sardor wrote: $ P(x,0) \implies f(x^2)=f(x)^2 $, thus $ f(x)=f(-x) $ or $ f(x)=-f(-x) $. If $ f $ is odd, the $ f(2x^2)=0 $, so $ f(x)=0 $ for all $ x \in R_0 $ and since $ f $ is odd, we get $ f(x)=0 $, for any $ x $. Let $ f $ is even, then ... I don't think this line is valid. Even though $ f(x)=f(-x) $ or $ f(x)=-f(-x) $ at any particular x, that doesn't mean $f(x)=f(-x)$ over all x or $f(x)=-f(-x)$ over all x, like you seem to assume in the next few lines.

Arshia.esl

15.04.2021 10:49

Anyone???

pco

23.04.2021 14:58

gobathegreat wrote: Find all functions $ f:\mathbb{R}\rightarrow\mathbb{R} $ such that for all $x,y\in{{\mathbb{R}}}$ holds $f(x^2)+f(2y^2)=(f(x+y)+f(y))(f(x-y)+f(y))$ Let $P(x,y)$ be the assertion $f(x^2)+f(2y^2)=(f(x+y)+f(y))(f(x-y)+f(y))$ Let $a=f(0)$ Subtracting $P(0,-x)$ from $P(0,x)$, we get $f(x)^2=f(-x)^2$ And so $\forall x$, either $f(-x)=f(x)$, either $f(-x)=-f(x)$ $P(0,0)$ $\implies$ $a=2a^2$ and so $a=0$ or $a=\frac 12$ $P(0,x)$ $\implies$ $f(2x^2)=4f(x)^2-a$ $P(x,0)$ $\implies$ $f(x^2)=f(x)^2+2af(x)+a^2-a$ $P(x,x)$ $\implies$ $(f(x)+a)(f(2x)-4f(x)+3a)=0$ 1) If $a=\frac 12$, then $\boxed{\text{S1 : }f(x)=\frac 12\quad\forall x}$

2) $a=0$ 2.1) Some basic stuff $P(0,x)$ $\implies$ $f(2x^2)=4f(x)^2$ $P(x,0)$ $\implies$ $f(x^2)=f(x)^2$ (and so $f(x)\ge 0\quad\forall x\ge 0$) $P(x,x)$ $\implies$ $f(x)(f(2x)-4f(x))=0$ If $f(x)\ne 0$ : $P(0,x)$ $\implies$ $4f(x)^2=2f(x)(f(-x)+f(x))$ and so $f(-x)=f(x)$ If $f(x)=0$ : $P(-x,x)$ $\implies$ $f(-x)^2=0$ and so $f(-x)=f(x)$ again. So $f(-x)=f(x)$ $\forall x$ and $f(x)$ is an even function. (and so $f(x)\ge 0\quad\forall x$) $f(2x^2)=4f(x^2)$ and so $f(2x)=4f(x)$ $\forall x\ge 0$ and, since even, $f(2x)=4f(x)\quad\forall x$ 2.2) Study of $U=f^{-1}(\{0\})$ : $U\in\{\{0\},\mathbb R\}$

Let $U=f^{-1}(\{0\})$. $0\in U$ 2.2.1) $u,v\in U$ implies $-u,2u,u^2,\frac u2,u+v,u-v\in U$ Proof"basic stuff" paragraph shows that $u\in U$ implies $-u,2u,u^2,\frac u2\in U$ Let $u,v\in U$: $P(u,v)$ $\implies$ $f(u+v)f(u-v)=0$ If $f(u+v)=0$, then $P(u-v,u+v)$ $\implies$ $f(u-v)^2=f(2u)f(-2v)=0$ If $f(u-v)=0$, then $P(u+v,u-v)$ $\implies$ $f(u+v)^2=f(2u)f(2v)=0$ And so $u,v\in U$ $\implies$ $u+v,u-v\in U$ 2.2.2) $f(x+u)=f(x)$ $\forall x$ and $\forall u\in U$ ProofLet $x\in\mathbb R$ and $u\in U$ If $x\in U$, previous result implies $f(x+u)=f(x-u)=f(x)=0$ If $x\notin U$, and so $f(x)\ne 0$ : $P(x,u)$ $\implies$ $f(x)^2=f(x+u)f(x-u)$ $P(u,x)$ $\implies$ $3f(x)^2=f(x+u)f(x-u)+f(x)(f(x+u)+f(x-u))$ Subtracting, we get $f(x+u)+f(x-u)=2f(x)$ and since we also have (two lines above) $f(x+u)f(x-u)=f(x)^2$, We get $f(x+u)=f(x-u)=f(x)$ Q.E.D. 2.2.3) If $f(x)=f(y)$, then either $x+y\in U$, either $x-y\in U$ (either both) likely improvable proofIf $f(x)=f(y)\ne 0$ $P(x,y)$ $\implies$ $4f(x)^2=f(x+y)f(x-y)+f(x)(f(x+y)+f(x-y)$ $P(x+y,x)$ $\implies$ $f(x+y)^2+2f(x)^2=2f(x)f(2x+y)$ $P(x,x+y)$ $\implies$ $f(x)^2+3f(x+y)^2=f(2x+y)(f(x)+f(x+y))+f(x)f(x+y)$ Cancelling $f(2x+y)$ between these two lines, we get $f(x+y)(f(x+y)-f(x))(f(x+y)-4f(x))=0$ And so $f(x+y)\in\{0,f(x),f(2x)\}$ Note that similartly, we have $f(x)=f(-y)$ and so $f(x-y)\in\{0,f(x),f(2x)\}$ If $f(x+y)=f(x)$, $P(x,y)$ $\implies$ $f(x-y)=\frac 32f(x)$, impossible since $f(x)\ne 0$ and $f(x-y)\in\{0,f(x),f(2x)\}$ So $f(x+y)\in\{0,f(2x)\}$ And, similartly $f(x-y)\in\{0,f(2x)\}$ If $f(x+y)$ and $f(x-y)$ are nonzero, this implies $f(x+y)=f(x-y)=f(2x)\ne 0$ Applying then the previous beginning of this paragraph to $(x+y,x-y)$ instead of $(x,y)$, we get : Both $f(2x),f(2y)\in\{0,f(2x+2y)\}$ and so $f(2x)=f(2y)=f(2x+2y)$, in contradiction with $f(x+y)=f(2x)$ So, either $f(x+y)=0$, either $f(x-y)=0$, still true when $f(x)=f(y)=0$ Q.E.D. 2.2.4) Either $U=\{0\}$, either $U=\mathbb R$ ProofLet $x\in\mathbb R$ and $u\in\mathbb R$. Note that $u^2\in U$. $f(x^2+2ux)=f(x^2+2ux+u^2)=f((x+u)^2)=f(x+u)^2=f(x)^2=f(x^2)$ and so either $x^2+ux\in U$, either $ux\in U$ $f(x^2-2ux)=f(x^2-2ux+u^2)=f((x-u)^2)=f(x-u)^2=f(x)^2=f(x^2)$ and so either $x^2-ux\in U$, either $ux\in U$ So, if $ux\notin U$, then both $x^2+ux,x^2-ux\in U$, and so their difference too, and contradiction So $ux\in U$ $\forall x$ And so, if $\exists u\in U\setminus\{0\}$ : $u\frac xu=x\in U$ $\forall x$ Q.E.D.

2.3) If $U=\mathbb R$ : $\boxed{\text{S2 : }f(x)=0\quad\forall x}$, which indeed fits Trivial. 2.4) If $U=\{0\}$ : $\boxed{\text{S3 : }f(x)=x^2\quad\forall x}$, which indeed fits

So we have $f(x)=0$ $\iff$ $x=0$ And $f(x)=f(y)$ $\iff$ $x=\pm y$ (remember either $x+y\in U$, either $x-y\in U$) Let $x,y$ such that $x\ne\pm y$ (and so $f(x)\ne f(y)$ $P(x,y)$ becomes $f(x)^2+3f(y)^2=f(x+y)f(x-y) + f(y)f(x+y)+f(y)f(x-y)$ Swapping there $x,y$ and subtracting, we get New assertion $Q(x,y)$ : $f(x+y)+f(x-y)=2f(x)+2f(y)$ $\forall x\ne \pm y$, still true when $x=\pm y$ Simple induction gives $f(x+ny)=n^2f(y)+n(f(x+y)-f(x)-f(y))+f(x)$ $\forall x,y$ and $\forall n\in\mathbb Z$ (note that this implies $f(nx)=n^2f(x)$) $n=2$ here above is $f(x+2y)=2f(x+y)+2f(y)-f(x)$ Plugging this in $P(x+y,y)$, we get : $f(x+y)^2-2(f(x)+f(y))+(f(x)-f(y))^2=0$ Which is a quadratic with two roots : $f(x)+f(y)\pm 2\sqrt{f(x)f(y)}$ And so : $\forall x,y$ : Either $f(x+y)=f(x)+f(y)+2\sqrt{f(x)f(y)}$ Either $f(x+y)=f(x)+f(y)-2\sqrt{f(x)f(y)}$ Note that we also have : Either $f(x-y)=f(x)+f(y)+2\sqrt{f(x)f(y)}$ Either $f(x-y)=f(x)+f(y)-2\sqrt{f(x)f(y)}$ And since, if $x,y\ne 0$, we have $f(x+y)\ne f(x-y)$, we get : $\forall x,y\ne 0$ : Either $f(x+y)=f(x)+f(y)+2\sqrt{f(x)f(y)}$ and $f(x-y)=f(x)+f(y)-2\sqrt{f(x)f(y)}$ Either $f(x+y)=f(x)+f(y)-2\sqrt{f(x)f(y)}$ and $f(x-y)=f(x)+f(y)+2\sqrt{f(x)f(y)}$ And so $f(x+y)^2+f(x-y)^2=2(f(x)+f(y))^2+8f(x)f(y)$ $\forall x,y\ne 0$, still true when $x=0$ or $y=0$ But $f(x+y)^2+f(x-y)^2=f((x+y)^2)+f((x-y)^2)=f(x^2+y^2+2xy)+f(x^2+y^2-2xy)$ And, using $Q(x^2+y^2,2xy)$, this is $2f(x^2+y^2)+2f(2xy)$ And so $f(x^2+y^2)+f(2xy)=(f(x)+f(y))^2+4f(x)f(y)$ So : If $f(x^2+y^2)=f(x^2)+f(y^2)+2f(x)f(y)$, previous line gives $f(xy)=f(x)f(y)$ If $f(x^2+y^2)=f(x^2)+f(y^2)-2f(x)f(y)$, previous line gives $f(xy)=2f(x)f(y)$ So, $\forall x,y$ : Either $f(x^2+y^2)=f(x^2)+f(y^2)+2f(x)f(y)$ and $f(xy)=f(x)f(y)$ Either $f(x^2+y^2)=f(x^2)+f(y^2)-2f(x)f(y)$ and $f(xy)=2f(x)f(y)$ But $f(xy)=2f(x)f(y)$ would imply $f(xy)^2=4f(x)^2f(y)^2$ and so $f(x^2y^2)=4f(x^2)f(y^2)$, impossible if $x,y\ne 0$ So $f(x^2+y^2)=f(x^2)+f(y^2)+2f(x)f(y)$ and $f(xy)=f(x)f(y)$ $\forall x,y\ne 0$, still true when $x=0$ or $y=0$ Setting then $g(x)=\sqrt{f(x)}\ge 0$, previous line gives $g(x^2+y^2)=g(x^2)+g(y^2)$ and $g(xy)=g(x)g(y)$ So $g(x)$ is additive, multiplicative and positive over $\mathbb R^+$ and so is $g(x)=x\quad\forall x\ge 0$ So $f(x)=x^2\quad\forall x\ge 0$ And so, since even, $f(x)=x^2\quad\forall x$ which indeed fits (Sophie-Germain identity). Q.E.D.

electrovector

29.07.2022 20:36

pco wrote: $P(0,x)$ $\implies$ $f(2x^2)=4f(x)^2-a$ This part is wrong since we cannot exactly say either $f(x)+f(-x)=0$ or $f(x)=f(-x)$. pco wrote: 1) If $a=\frac 12$, then $\boxed{\text{S1 : }f(x)=\frac 12\quad\forall x}$ However realizing $f(x)^2=f(-x)^2$ and $f((-x)^2)=f(x^2)$ gives $f(x)=f(-x)$ because of $P(x, 0)$ and $a \neq 0$.

ZETA_in_olympiad

01.09.2022 20:59

Clearly one of the hardest functional equations to appear in a test. For another solution, one may look at the official solution , it is much like the solution in #7, but a bit slower.