## Search found 40 matches

### Re: Equality and square

The case of negative integers is quite trivial. Now we'll work with case $a,b > 0$ Lemma 1: $x < 2^x$ Proof: We'll prove it by induction. Base case is solved. Now assume $x<2^x \Rightarrow x+1<2^x+1 \le 2^{x+1}$. Done! Lemma 2: $b^2 < 2^{2b+1}$ Proof: We'll prove it by induction. Base case is solved...
Sat Aug 05, 2017 11:54 pm

Forum: Number Theory
Topic: Equality and square
Replies: 2
Views: 219

NO INJECTIVITY!!! Notice that, if $f(x)$ is a solution, then $-f(x)$ is also a solution. Let $P(x,y)$ denotes the assertion. $P(0,0)$ implies $f(f(0)^2)=0$ . Now if $f(0)=0$ , $P(x,0)$ implies $f(x)=0$ for all $x \in \mathbb{R} ... Thu Jul 27, 2017 12:07 am Forum: International Mathematical Olympiad (IMO) Topic: IMO 2017 Problem 2 Replies: 3 Views: 654 ### Re: Beginner's Marathon An easy problem:$\text{Problem 20}$The fractions$\frac{7x+1}{2}$,$\frac{7x+2}{3}$,.......,$\frac{7x+2016}{2017}$are irreducible.Find all possible values of$x$such that$x$is less than or equal to$300$.This Problem was recommended by Zawad bhai in a mock test at CMC. Here's a hint for you g... Sun Jun 11, 2017 12:32 am Forum: Junior Level Topic: Beginner's Marathon Replies: 68 Views: 3371 ### Re: IMO 2001 Problem1 Our problem is equivalent to$OP>CP$. Lemma 1:$BP \ge R+PC$where$R$is the circumradius of$\triangle ABC$. Proof of lemma:$BP-PC=AB \cos \angle B - AC \cos \angle C = 2R(\sin \angle C \cos \angle B - \sin \angle B \cos \angle C ) \ge R$because$\angle C- \angle B \ge 30$. Now we back ... Mon Jun 05, 2017 12:02 am Forum: International Mathematical Olympiad (IMO) Topic: IMO 2001 Problem1 Replies: 2 Views: 130 ### Re: Particular divisibility.. Case 1:$a=b>1a^2-a|a^2+a$or$a^2-a|2a$. So,$a \le 3$. Checking the values, we get$a=2$and$a=3$. Case 2:$a \neq b$WLOG$a>ba^2-b|b^2+a \Rightarrow b^2+a \ge a^2-b \Rightarrow b(b+1) \ge a(a-1)$. As$a>b$, we get$b+1>a-1 \Rightarrow a-b<2 \Rightarrow a=b+1$.$b^2-a|a^2+...
Sat Jun 03, 2017 11:37 pm

Forum: Number Theory
Topic: Particular divisibility..
Replies: 1
Views: 160

$M$ is the midpoint of $AC$ and $X$ is the midpoint of $AF$. Proving $\angle DEF = 90$ is equivalent to $MDEF$ cyclic or $\angle MEF = \angle MDF = 30$ It is easy to show $\triangle DFX$ is equilateral. So, $ABDX$ is cyclic. $\angle EFC = \angle BXC = 60+\angle BAD$, so $\angle FEC = 90 - \angle BA... Sat Jun 03, 2017 8:54 pm Forum: Geometry Topic: CGMO 2007/5 Replies: 1 Views: 120 ### Re: A strange divisibility$\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...
Fri Jun 02, 2017 6:48 am

Forum: Number Theory
Topic: A strange divisibility
Replies: 1
Views: 208

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 $ABCP$,$... Sat May 20, 2017 12:45 am Forum: Geometry Topic: ISL 2006 G3 Replies: 2 Views: 163 ### Re: Inequality with abc = 1 My solution. Same as Asif Bhai's.$(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}) ...
Sat May 20, 2017 12:07 am

Forum: Algebra
Topic: Inequality with abc = 1
Replies: 3
Views: 189

### 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&...
Thu May 11, 2017 10:37 pm

Forum: Number Theory
Topic: Divisibility with a and b
Replies: 2
Views: 166
Next