[tex]Conditia~de~existenta~a~radicalului~este~40 \geq 4 \sqrt{n-3} \Leftrightarrow \\ \\ \Leftrightarrow 10 \geq \sqrt{n-3} \Leftrightarrow 100 \geq n-3 \Leftrightarrow 103 \geq n~si~n \geq 3. \\ \\ \sqrt{40-4 \sqrt{n-3}}=k \in N \Rightarrow 40-4 \sqrt{n-3}}=k^2 \Rightarrow \sqrt{n-3}= \frac{40-k^2}{4} \in Q [/tex]
[tex]Folosind~proprietatea~"Daca~m \in N,~iar ~\sqrt{m} \in Q,~atunci~\sqrt{m} \in N", \\ \\ deducem~ca~ \sqrt{n-3} \in N \Rightarrow n \in \{3,4,7,12,19,28,39,52,67,84 ,103 \}. \\ \\ Proband~fiecare~valoare~se~constata~ca~numerele~care~verifica~ \\ \\ cerinta~problemei~sunt:~n \in \{4,39,84\}\\ \\ m_a= \frac{4+39+84}{3}= 42,(3).[/tex]
