IMO Shortlist 2012 N1
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
We call a set $A$ of integers admissible if $x,y\in A$(not necessarily distinct), then $x^2+kxy+y^2\in A$ for every integer $k$.
Determine all pairs of nonzero integers $m,n$ for which if $m,n\in A$, then $A=\mathbb{Z}$.
Determine all pairs of nonzero integers $m,n$ for which if $m,n\in A$, then $A=\mathbb{Z}$.
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: IMO Shortlist 2012 N1
Hint:
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.
- nahin munkar
- Posts:81
- Joined:Mon Aug 17, 2015 6:51 pm
- Location:banasree,dhaka
Re: IMO Shortlist 2012 N1
OK. I have approached it like this.
Let, $d= gcd(m,n).$ If, $d > 1$, if $d|x$ & $ d|y $ then $d|x^2+kxy+y^2 $ for all k.So,though the multiple of d satisfies condition, but $A \ne \mathbb{Z}$. here it is easy to see, $ d=1$ .
Now,We let,the set $A$ is admissible containing m,n.So,if $ x^2 \in A$ , then,$ kx^2 \in A$ (calculating from condition). if $ d=1. gcd(m^2,n^2)=1$ . According to hint ,we get $ am^2+bn^2=1$ by bezout's theorem. We get here $(am^2+bn^2)^2=1 \in A...(*)$, where k =2. & also $am^2 \in A, bn^2\in A.$ From (*), $ 1 \in A $ ; As, all multiples of $ m^2 \in A, so, k*1^2=k \in A.$ For this, all integers $ \in A.$ $ SO, A=\mathbb{Z} . $.......
Let, $d= gcd(m,n).$ If, $d > 1$, if $d|x$ & $ d|y $ then $d|x^2+kxy+y^2 $ for all k.So,though the multiple of d satisfies condition, but $A \ne \mathbb{Z}$. here it is easy to see, $ d=1$ .
Now,We let,the set $A$ is admissible containing m,n.So,if $ x^2 \in A$ , then,$ kx^2 \in A$ (calculating from condition). if $ d=1. gcd(m^2,n^2)=1$ . According to hint ,we get $ am^2+bn^2=1$ by bezout's theorem. We get here $(am^2+bn^2)^2=1 \in A...(*)$, where k =2. & also $am^2 \in A, bn^2\in A.$ From (*), $ 1 \in A $ ; As, all multiples of $ m^2 \in A, so, k*1^2=k \in A.$ For this, all integers $ \in A.$ $ SO, A=\mathbb{Z} . $.......
Last edited by nahin munkar on Sun Aug 07, 2016 7:21 pm, edited 1 time in total.
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss
- asif e elahi
- Posts:185
- Joined:Mon Aug 05, 2013 12:36 pm
- Location:Sylhet,Bangladesh
Re: IMO Shortlist 2012 N1
How $am^2 \in A$?nahin munkar wrote:OK. I have approached it like this.
Let, $d= gcd(m,n).$ If, $d > 1$, if $d|x$ & $ d|y $ then $d|x^2+kxy+y^2 $ for all k.So,though the multiple of d satisfies condition, but $A \ne \mathbb{Z}$. here it is easy to see, $ d=1$ .
Now,We let,the set $A$ is admissible containing m,n.So,if $ x^2 \in A$ , then,$ kx^2 \in A$ (from condition). if $ d=1. gcd(m^2,n^2)=1$ . According to hint ,we get $ am^2+bn^2=1$ by bezout's theorem. We get here $(am^2+bn^2)^2=1 \in A...(*)$, where k =2. & also $am^2 \in A, bn^2\in A.$ From (*), $ 1 \in A $ ; As, all multiples of $ m^2 \in A, so, k*1^2=k \in A.$ For this, all integers $ \in A.$ $ SO, A=\mathbb{Z} . $.......
- nahin munkar
- Posts:81
- Joined:Mon Aug 17, 2015 6:51 pm
- Location:banasree,dhaka
Re: IMO Shortlist 2012 N1
asif e elahi wrote:How $am^2 \in A$?nahin munkar wrote:OK. I have approached it like this.
Let, $d= gcd(m,n).$ If, $d > 1$, if $d|x$ & $ d|y $ then $d|x^2+kxy+y^2 $ for all k.So,though the multiple of d satisfies condition, but $A \ne \mathbb{Z}$. here it is easy to see, $ d=1$ .
Now,We let,the set $A$ is admissible containing m,n.So,if $ x^2 \in A$ , then,$ kx^2 \in A$ (from condition). if $ d=1. gcd(m^2,n^2)=1$ . According to hint ,we get $ am^2+bn^2=1$ by bezout's theorem. We get here $(am^2+bn^2)^2=1 \in A...(*)$, where k =2. & also $am^2 \in A, bn^2\in A.$ From (*), $ 1 \in A $ ; As, all multiples of $ m^2 \in A, so, k*1^2=k \in A.$ For this, all integers $ \in A.$ $ SO, A=\mathbb{Z} . $.......
As, $ kx^2 \in A $ for all $ k $ if $ x \in A $ (calculation from condition). Here, $ m \in A $ . $ So $, $ am^2 \in A $.
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss
- asif e elahi
- Posts:185
- Joined:Mon Aug 05, 2013 12:36 pm
- Location:Sylhet,Bangladesh
Re: IMO Shortlist 2012 N1
There is no condition saying $kx^2 \in A$. You have to prove it. (Though the proof is very obvious).
- Phlembac Adib Hasan
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Re: IMO Shortlist 2012 N1
Interesting sidenote: if I recall correctly, ISL 2012 A2 is somewhat similar in flavor.
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- nahin munkar
- Posts:81
- Joined:Mon Aug 17, 2015 6:51 pm
- Location:banasree,dhaka
Re: IMO Shortlist 2012 N1
$ OK $. if we let $ x=y $ of the condition. Then, a simple manipulation tells us that it is obvious $ (kx^2 \in A) $ for all $ k $ under admissible-set condition . Plugging, $ x=y $ of the defination, we now easily get the proof......asif e elahi wrote:There is no condition saying $kx^2 \in A$. You have to prove it. (Though the proof is very obvious).
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss