Search found 4 matches
- Wed Sep 28, 2022 2:47 pm
- Forum: Algebra
- Topic: FE Marathon!
- Replies: 98
- Views: 658359
Re: FE Marathon!
Let $P(x,y)$ denote the original equation. $P(0,0)\Rightarrow f(0)=0.$ $P(-x,y)\Rightarrow f(x)=-f(x) \Rightarrow f \text{ is odd.}$ WLOG $x,y \geq 0.$ $P(x,0)\Rightarrow f(x^2)=xf(x) \Rightarrow f(x^2-y^2)=f(x^2)-f(y^2)$ $\Rightarrow f(x^2)=f(x^2-y^2)+f(y^2).$ $\exists z\geq 0: z^2=x \Rightarrow f(...
- Tue May 10, 2022 3:48 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2011 Problem 3
- Replies: 2
- Views: 9595
Re: IMO 2011 Problem 3
Let $P(x,y)$ be the given assertion. Comparing $P(x,f(y)-x)$ and $P(y,f(x)-y)$ yields, $$xf(x)+yf(y)\leq 2f(x)f(y).$$ $y\mapsto 2f(x)\Rightarrow xf(x)\leq 0. \qquad (*)$ --------------------- $\textbf{Claim: }f(k)\leq 0~~\forall k.$ $Proof.$Suppose $\exists k:f(k)>0,$ then $$f(k+y)\leq yf(k)+f(f(k))...
- Sat Mar 19, 2022 3:07 pm
- Forum: National Math Camp
- Topic: Problem - 02 - National Math Camp 2021 Number Theory Exam - "Group theory"
- Replies: 3
- Views: 10153
- Fri Mar 18, 2022 2:29 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2017 Problem 2
- Replies: 4
- Views: 14987
Re: IMO 2017 Problem 2
]Let $P(x,y)$ denote the assertion $f(f(x)f(y))+f(x+y)=f(xy).$ If $x \neq 1$, $P(x,\frac{x}{x-1}) \implies f(f(x)f(\frac{x}{x-1}))=0 \text{ [eq.1]} \implies f(0)=0$. Claim 1: $f(x)=0$ Proof. $\exists r: f(r)f(\frac{r}{r-1})\neq 1 \forall r \in \mathbb{R}-\{1\}$. Setting $x=f(r)f(\frac{r}{r-1})$ mean...