IMO Shortlist 2012 N1

[Thanic Nur Samin]

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 $A=\mathbb{Z}$.

[Thanic Nur Samin]

Hint:

Use Bézout's identity to show $1\in A$ for coprime $m,n$.

[nahin munkar]

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} . $.......

[asif e elahi]

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} . $.......

How $am^2 \in A$?

[nahin munkar]

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} . $.......

How $am^2 \in A$?

As, $ kx^2 \in A $ for all $ k $ if $ x \in A $ (calculation from condition). Here, $ m \in A $ . $ So $, $ am^2 \in A $.

[asif e elahi]

There is no condition saying $kx^2 \in A$. You have to prove it. (Though the proof is very obvious).

[Phlembac Adib Hasan]

Interesting sidenote: if I recall correctly, ISL 2012 A2 is somewhat similar in flavor.

[nahin munkar]

There is no condition saying $kx^2 \in A$. You have to prove it. (Though the proof is very obvious).

$ 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......