Question

Aratati ca numarul A= $2n^{2} + [ \sqrt{4n ^{2}+n } ]+1$,unde n este numar natural nenul , poate fi scris ca o suma de 2 patrate perfecte.(Am notat [a] partea intreaga)

Answer 1

$n \in N^* \Rightarrow 4n^2\ \textless \ 4n^2+n\ \textless \ 4n^2+4n+1 \Leftrightarrow \\ \\ \Leftrightarrow 4n^2\ \textless \ 4n^2+n\ \textless \ (2n+1)^2 \Rightarrow 2n\ \textless \ \sqrt{4n^2+n}\ \textless \ 2n+1 \Rightarrow \\ \\ \Rightarrow [\sqrt{4n^2+n}]=2n. \\ \\ A=2n^2+2n+1=n^2+(n^2+2n+1)=n^2+(n+1)^2.$