## Search found 64 matches

Fri Jun 02, 2017 6:48 am
Forum: Number Theory
Topic: A strange divisibility
Replies: 1
Views: 1297

$\frac {x^2}{2xy^2-y^3+1}=k \in \mathbb{N}$ If $y=1$ , we can easily find $x=2k$ . Now, let $y>1$. Then, $\frac {x^2}{2xy^2-y^3+1}=k$ gives us the quadratic equation $x^2-2ky^2x+k(y^3-1)=0$ . Here the discriminant $X=4k^2y^4-4ky^3+4k$ is a perfect square . Notice that $(2ky^2-y-1)^2 < X < (2ky^2-y+... Sat May 20, 2017 12:45 am Forum: Geometry Topic: ISL 2006 G3 Replies: 2 Views: 7697 ### Re: ISL 2006 G3 Solution with spiral similarity is easy. Here I'm gonna give a different solu. And I chased some angles and lengths.$\triangle AED$and$\triangle ABC$are similar and so$AC.AD=AB.AE$which along with$\angle BAD = \angle CAE$implies$\triangle ABD$and$\triangle ACE$are similar and so$ABCP$,$...
Sat May 20, 2017 12:07 am
Forum: Algebra
Topic: Inequality with abc = 1
Replies: 3
Views: 4873

My solution. Same as Asif Bhai's. $(x-1+\frac{1}{y})(y-1+\frac{1}{z})=xy-x+\frac{x}{z}-y+1-\frac{1}{z}+1-\frac{1}{y}+\frac{1}{yz}=\frac{x}{z}+2-(y+\frac{1}{y})\le \frac{x}{z}$ because $y+\frac{1}{y} \ge 2$ Similarly we get, $(y-1+\frac{1}{z})(z-1+\frac{1}{x}) \le \frac{y}{x}$ and $(z-1+\frac{1}{x})(... Thu May 11, 2017 10:37 pm Forum: Number Theory Topic: Divisibility with a and b Replies: 2 Views: 1544 ### Re: Divisibility with a and b$\frac{b^3-1}{ab-1}=x$Firstly notice that if$(a,b)$is a solution, then$(b,a)$is also a solution. So, we may assume wlog$a \ge b$. If$b=1$,$x=0$. So, we got a solution$(a,1)$where$a>1$. Now consider the case$b \ge 2$.$x>0$. Notice that$x \equiv 1 (mod b)$. So,$x=nb+1$. As$a \ge...
Thu May 04, 2017 8:05 pm
Forum: Algebra
Topic: FE ^_^
Replies: 1
Views: 3917

### Re: FE ^_^

Thu May 04, 2017 7:59 pm
Forum: Algebra
Topic: FE ^_^
Replies: 1
Views: 3917

### FE ^_^

Find all $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for all distinct $x,y,z$
$f(x)^2-f(y)f(z)=f(x^y)f(y)f(z)[f(y^z)-f(z^x)]$
Mon May 01, 2017 9:45 pm
Forum: Number Theory
Topic: IMO Shortlist 2011 N5
Replies: 4
Views: 2526

### Re: IMO Shortlist 2011 N5

$f(m-n)|f(m)-f(n)$ $n=0$ implies $f(m)|f(0)$ $m=0$ and $n=m$ implies $f(-m)|f(m)$ . Similarly we get $f(m)|f(-m)$ . So, $f(m)=f(-m)$ $n=-n$ implies $f(m+n)|f(m)-f(n)$ from which we get $f(m+n) \le f(n)-f(m)$ Again, $m=m+n$ and $n=m$ implies $f(n)|f(m+n)-f(m)$ . Assume $f(m) \neq f(m+n)$ . Here we'...
Thu Apr 27, 2017 12:12 am
Forum: Junior Level
Topic: 14x14 grid
Replies: 1
Views: 1480

### Re: 14x14 grid

Put $1$ and $0$ alternately in that grid. Then in any $T$ shape, $2$ configuration is possible: #1 three $1$'s and one $0$'s #2 three $0$'s and one $1$'s In both configuration, the sum of the numbers in that $T$ is odd. The grid is $14 x 14$. So, we need an odd number of $T$ - tiles. The initial su...
Wed Apr 26, 2017 12:13 am
Forum: Number Theory
Topic: JBMO:NT
Replies: 1
Views: 1286