Question

Dau 50 de puncte celui care rezolva problema.

Answer 1

$\\ Se cer numerele naturale \overset{-}{ab} \Rightarrow a,b \in \mathbb_{N} \\ \\\dfrac{9}{a^2+b^2}\in \mathbb_{N} \\ \\ \Rightarrow a^2+b^2\in D_9^+ \Rightarrow a^2+b^2 \in \Big\{1,3,9\Big\} \\ \\ \boxed{1}\quad a^2+b^2=1 \Rightarrow a = 0 si b = 1 \quad sau \quad a = 1 si b = 0 \Rightarrow \\ \\ \Rightarrow (a,b) = \Big\{(0,1);(1,0)\Big\}\Rightarrow \overset{-}{ab} = \Big\{01;10\Big\}, dar , 01 nu exista \Rightarrow \\ \\ \Rightarrow \overset{-}{ab} = \Big\{10\Big\}$

$\boxed{2} \quad a^2+b^2 = 3 \quad (nu exista nicio posibilitate pentru a,b\in\mathbb_{N} ) \\ \\ \boxed{3}\quad a^2+b^2 = 9 \Rightarrow a = 0 si b = 3 \quad sau \quad a = 3 si b = 0 \Rightarrow \\ \\ \Rightarrow (a,b) = \Big\{(0,3);(3,0)\Big\}\Rightarrow \overset{-}{ab} = \Big\{03;30\Big\}, dar , 03 nu exista \Rightarrow \\ \\ \Rightarrow \overset{-}{ab} = \Big\{30\Big\}$

$Din \boxed{1} \cup \boxed{2}\cup\boxed{3} \Rightarrow \boxed{\overset{-}{ab} = \Big\{10;30\Big\}}$