Search found 46 matches
- Thu Mar 11, 2021 10:53 pm
- Forum: Higher Secondary Level
- Topic: A subtle discussion of a prob from regional 2019
- Replies: 7
- Views: 8004
Re: A subtle discussion of a prob from regional 2019
The question was, $f(x+y)=xf(x)+2(y)-x-y$ $f(2019)=\frac{a}{b} ; a+b = ?$ If we use $x=0$ and $y=2019$ we get that $f(2019)=2019$ but if we use the opposite where $x = 2019$, and $y=0$ it gets messier[Probably gives something definitly not 2019], was my first idea correct or we got work to do? And ...
- Tue Mar 02, 2021 10:06 am
- Forum: Algebra
- Topic: FE Marathon!
- Replies: 98
- Views: 655902
Re: FE Marathon!
$\textbf{Problem 17}$ Find all functions $f: \mathbb R \to \mathbb R$ such that\[ f( xf(x) + f(y) ) = f(x)^2 + y \]for all $x,y\in \mathbb R$. DO MENTION THE SOURCE :( :( Proof(Not sure this the final version :? ): It is easy to see the $f(x)$ is bijectve just fix any $x=x_0$ and the rest is easy. ...
- Sat Feb 27, 2021 8:08 pm
- Forum: Algebra
- Topic: FE Marathon!
- Replies: 98
- Views: 655902
- Sat Feb 27, 2021 7:02 pm
- Forum: Algebra
- Topic: FE Marathon!
- Replies: 98
- Views: 655902
Re: FE Marathon!
$\textbf{Problem 17}$
Find all functions $f: \mathbb R \to \mathbb R$ such that\[ f( xf(x) + f(y) ) = f(x)^2 + y \]for all $x,y\in \mathbb R$.
Find all functions $f: \mathbb R \to \mathbb R$ such that\[ f( xf(x) + f(y) ) = f(x)^2 + y \]for all $x,y\in \mathbb R$.
- Sat Feb 27, 2021 6:36 pm
- Forum: Algebra
- Topic: FE Marathon!
- Replies: 98
- Views: 655902
Re: FE Marathon!
$Problem 16$ Find all function such that $f:R \rightarrow R$ and $f(xf(y)+y)+f(xy+x)=f(x+y)+2xy$ $\textbf{Solution 16}$ Let $P(x,y)$ be the assertion. $P(1,\frac{x}{2})\Rightarrow f(f(\frac{x}{2})+\frac{x}{2})=x \Rightarrow f$ is surjective $\Rightarrow$ there exists $a$ such that $f(a)=1$. $P(\fra...
- Sat Feb 27, 2021 2:51 pm
- Forum: Number Theory
- Topic: Existence of a prime factor>p
- Replies: 4
- Views: 6279
Re: Existence of a prime factor>p
Prove that if $p$ is a prime, then $p^p-1$ has a prime factor, greater than $p.$ :D $p^p-1=(p-1)(p^{p-1}+p^{p-2}+\cdots+1)$ Now let $q$ be a prime factor of $p^p-1$ such that $q\nmid p-1$. This prime divisor obviously exists because if a prime factor of $p^p-1$, let it be $r$,divides $p-1$, then we...
- Sat Feb 27, 2021 9:44 am
- Forum: Algebra
- Topic: FE Marathon!
- Replies: 98
- Views: 655902
Re: FE Marathon!
2004 IMO SLAsif Hossain wrote: ↑Sat Feb 27, 2021 8:59 amPlease do mention the source (i think i have seen it in the shortlist)
- Fri Feb 26, 2021 11:44 am
- Forum: Algebra
- Topic: FE Marathon!
- Replies: 98
- Views: 655902
Re: FE Marathon!
$\textbf{Problem 15}$
Find all functions $ f: \mathbb{N}\to \mathbb{N}$ satisfying
\[ \left(f(m)^2+f\left(n\right)\right) \mid \left(m^{2}+n\right)^{2}\]for any two positive integers $ m$ and $ n$.
Find all functions $ f: \mathbb{N}\to \mathbb{N}$ satisfying
\[ \left(f(m)^2+f\left(n\right)\right) \mid \left(m^{2}+n\right)^{2}\]for any two positive integers $ m$ and $ n$.
- Fri Feb 26, 2021 11:13 am
- Forum: Algebra
- Topic: FE Marathon!
- Replies: 98
- Views: 655902
Re: FE Marathon!
Problem 14 : Find all functions $f:\mathbb{Z}\mapsto\mathbb{Z}$ that satisfies : $f(x-f(y))=f(f(x))-f(y)-1$ For all $x,y\in\mathbb{Z}$. Source : IMO Shortlist 2015, A2 $\textbf{Solution 14:}$ Let $P(x,y)$ be the assertion. $P(x,f(x))\Rightarrow f(x-f(f(x)))=-1$ $P(x,x-f(f(x)))\Rightarrow f(x+1)=f(f...
- Sun Feb 14, 2021 11:36 am
- Forum: Number Theory
- Topic: Existence of a prime factor>p
- Replies: 4
- Views: 6279
Re: Existence of a prime factor>p
Prove that if $p$ is a prime, then $p^p-1$ has a prime factor, greater than $p.$ :D $p^p-1=(p-1)(p^{p-1}+p^{p-2}+\cdots+1)$ Now let $q$ be a prime factor of $p^p-1$ such that $q\nmid p-1$. This prime divisor obviously exists because if a prime factor of $p^p-1$, let it be $r$,divides $p-1$, then we...