I don't have access to a pc currently, so just a sketch. Let $P(x,y)$ be the given equation. $P(x,0)$ implies $f(0)=0,f(1)=1$. $P(xy,y)-P(x,1)$ gives $f(xy)=f(x)f(y)$. So write $P(x,y)$ as $\frac {f(x+y)}{f(x-y)}=\frac {x+y}{x-y}$. Now let this be $Q(x,y)$. Use $P(2,1)P(3,1),P(4,1)P(5,1),f(6)=f(2)f(...

- Fri Feb 21, 2014 11:02 am
- Forum: Algebra
- Topic: FE from real to real
- Replies:
**2** - Views:
**457**

Yeah, and it directly kills SL-2006-N5. And this very problem(and a little গুতানি from Masum vai ) motivated me to learn about cyclotomic polynomials.

- Thu Feb 20, 2014 7:35 pm
- Forum: Number Theory
- Topic: prime divisors
- Replies:
**5** - Views:
**893**

By Zsigmondy's theorem $p^p-1$ has a prime divisor not dividing $p-1$[Check the exceptions, none works]. Let it be $q$. Let $d=ord_qp$. So $p^d \equiv 1 \pmod q$. So $d|p \Rightarrow d=1,p$. If $d=1$, then $q|p-1$, a contradiction! So $d=p$. Now $d|\phi (q) \Rightarrow p|q-1 \Rightarrow q=kp+1$, don...

- Tue Feb 18, 2014 9:30 pm
- Forum: Number Theory
- Topic: prime divisors
- Replies:
**5** - Views:
**893**

$[6]$( APMO'08 ) $\Gamma$ is the circum-circle of $\triangle ABC$. A circle passing through $A$ and $C$ meets $BC$ and $BA$ at $D$ and $E$ respectively. $AD$ and $CE$ intersect $\Gamma$ again at $G$ and $H$. The tangents of $\Gamma$ at $A$ and $C$ meet $DE$ at $L$ and $M$. Prove that, the intersect...

- Tue Feb 11, 2014 2:27 pm
- Forum: National Math Olympiad (BdMO)
- Topic: warm-up problems for national BdMO'14
- Replies:
**25** - Views:
**2256**

So let me get this straight $\mathbb B \rightarrow \mathbb B=\mathbb B, \mathbb B \rightarrow \mathbb W=\mathbb B, \mathbb W \rightarrow \mathbb B=\mathbb B, \mathbb W \rightarrow \mathbb W=\mathbb W$. WLOG we assume the square is rotated anti-clockwise. For simplicity, we assume the centre of a squ...

- Thu Feb 06, 2014 7:39 pm
- Forum: Combinatorics
- Topic: rotating a colored square
- Replies:
**5** - Views:
**1033**

Similar to a Balkan problem which was in 2012 TST too! Let $n^3=p^2-p-1 \Rightarrow (n+1)(n^2-n+1)=p(p-1)$. If $p|n+1$, then $n+1 \ge p \Rightarrow n(n+1) \ge p(p-1)=(n+1)(n^2-n+1) \Rightarrow n \ge n^2-n+1 \Rightarrow n=1,2.$ Plugging them in the original equation we find one solution $(n,p)=(1,2)$...

- Tue Feb 04, 2014 5:04 pm
- Forum: Number Theory
- Topic: positive cube = $p^2-p-1$
- Replies:
**7** - Views:
**730**

Mine involves duality as well. (Pascal is a result of dual projective axioms so far as I know :? ) Let $AC \cap BD=G, AC \cap PQ=G', AP \cap CQ=X, AQ \cap CP=Y$. Applying Pascal's theorem on $ADCQMP$, we get $AD \cap MQ=F, DC \cap MP=E, AP \cap CQ=X$ are collinear. Similarly applying Pascal's theore...

- Tue Jan 28, 2014 6:12 pm
- Forum: Geometry
- Topic: Geometry problem concerning duality
- Replies:
**2** - Views:
**682**

Labib wrote: I'm posting a solution posted by a brilliant member named Patrick Corn.

Lord pco

- Wed Jan 22, 2014 9:16 pm
- Forum: Junior Level
- Topic: MCQ highest
- Replies:
**3** - Views:
**800**

Am I missing something? Clearly $i \neq j$. So let $i<j$. Now by some algebraic manipulation we get \[\frac{\binom nj}{\binom ni}=\frac{\binom{n-i}{n-j}}{\binom ji}\] Since $LHS$ is irreducible, comparing denominators we have \[\binom ni|\binom ji \Rightarrow \binom ni \leq \binom ji\] which is a co...

- Tue Jan 07, 2014 5:53 pm
- Forum: Number Theory
- Topic: Nice In a Tricky Way!
- Replies:
**2** - Views:
**654**

King Jakaria wrote:Should the converse necessarily been true? That is, given ABC right-angled, should necessarily the center of A1B1C1 lie on the circumcircle of ABC?

Yes.

- Fri Nov 15, 2013 7:43 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2013, Day 1-P3
- Replies:
**3** - Views:
**1732**