Fie $a,b,c \in \Bbb{Z}$ astfel incat $a+b+c |a^2+b^2+c^2$. Demonstrati ca exista o infinitate de numere naturale $n$ astfel incat $a+b+c |a^n+b^n+c^n$.
Laurentiu Panaitopol
Test selectie Timisoara 1987
-
- Mesaje: 90
- Membru din: Mie Iul 21, 2010 11:37 pm
- Contact:
-
- Mesaje: 84
- Membru din: Dum Mar 27, 2011 10:36 pm
Re: Test selectie Timisoara 1987
Vom demonstra rezultatul prin inductie.
Notam intai prin $a^t+b^t+c^t=S_t$ si $a^tb^t+a^tc^t+b^tc^t=N_t$.
Lema: Daca$S_1 |S_n si S_1|S_{2n} \Rightarrow S_1 |S_{4n}$.
Demonstratie: $S_1|S_{2n} \Rightarrow S_1 | S_{4n} +2N_{2n}.$ Dar $S_1 | S_n \Rightarrow S_1 | S_{2n} + 2N_n \Rightarrow S_1 | 2N_n$ $\Rightarrow S_1 | 2N_{2n} + 4S_na^nb^nc^n \Rightarrow S_1 | 2N_{2n}$
Obtinem ca $S_1 | S_4n$.
Prin inductie, problema este rezolvata.
Notam intai prin $a^t+b^t+c^t=S_t$ si $a^tb^t+a^tc^t+b^tc^t=N_t$.
Lema: Daca$S_1 |S_n si S_1|S_{2n} \Rightarrow S_1 |S_{4n}$.
Demonstratie: $S_1|S_{2n} \Rightarrow S_1 | S_{4n} +2N_{2n}.$ Dar $S_1 | S_n \Rightarrow S_1 | S_{2n} + 2N_n \Rightarrow S_1 | 2N_n$ $\Rightarrow S_1 | 2N_{2n} + 4S_na^nb^nc^n \Rightarrow S_1 | 2N_{2n}$
Obtinem ca $S_1 | S_4n$.
Prin inductie, problema este rezolvata.