Točno
16. studenoga 2017. 02:32 (6 godine, 8 mjeseci)
Call admissible a set
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
of integers that has the following property:
If
![x,y \in A](/media/m/4/2/f/42fc8a4456132503d70c33e94401b687.png)
(possibly
![x=y](/media/m/c/6/6/c6628da2e638cae5d1afbabbd3cccf56.png)
) then
![x^2+kxy+y^2 \in A](/media/m/b/b/9/bb94a657146a463c87793202c5bb9d3a.png)
for every integer
![k](/media/m/f/1/3/f135be660b73381aa6bec048f0f79afc.png)
.
Determine all pairs
![m,n](/media/m/5/0/b/50b28d5a8b3362b2ac435400d318b3af.png)
of nonzero integers such that the only admissible set containing both
![m](/media/m/1/3/6/1361d4850444c055a8a322281f279b39.png)
and
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
is the set of all integers.
Proposed by Warut Suksompong, Thailand
%V0
Call admissible a set $A$ of integers that has the following property:
If $x,y \in A$ (possibly $x=y$) then $x^2+kxy+y^2 \in A$ for every integer $k$.
Determine all pairs $m,n$ of nonzero integers such that the only admissible set containing both $m$ and $n$ is the set of all integers.
Proposed by Warut Suksompong, Thailand
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Uočimo da ako imamo da je $m \in A $ tada iz uvijeta za $x=y=m$ imamo da je $m^2+kmm+m^2 \in A$ odnosno $ (k+2)m^2 \in A$, lijepše zapisano $$ lm^2\in A, \forall l\in \mathbb{Z}$$
Očito ako vrijedi $gcd(m,n)=1 \Rightarrow gcd(m^2,n^2)=1 $ te $\exists a,b \in \mathbb{Z}$ takvi da $$am^2+bn^2=1$$
Budući da su $m,n \in A$ imamo da su i $am^2,bn^2\in A \Rightarrow (am^2)^2+ 2am^2bn^2+(bn^2)^2 \in A$ odnosno $1=(am^2+bn^2)^2\in A$ te odabirom $x=y=1$ dobijamo $$k+2 \in A, \forall k \in \mathbb{Z}$$ odakle slijedi da je $A\equiv \mathbb{Z}$ za svake $m,n$ takve da je $gcd(m,n)=1$.
Teži dio zadatka je dokazati da za $m,n$ takve da je $gcd(m,n)=d>1$ vrijedi da $A \neq \mathbb{Z}$. Za to pronalazimo eksplicitnu konstrukciju admissible seta. Isprobavanjem dolazimo do zakljucka da je $$A=\{\enspace kd \enspace | \quad \forall k \in \mathbb{Z} \quad \}$$
Uočimo da su $m,n \in A$ te da za svaka dva broja $ad,bd \in A$ imamo da je i $(a^2d+kabd+b^2d)d \in A$. Tako $A$ zadovoljava uvjete admissible seta a ne popunjava skup svih cijelih brojeva.
16. studenoga 2017. 14:09 | abeker | Točno |