Question

Aratati ca exista n numar natural,astfel incat numarul n^2+n+41 sa fie patrat perfect(n^2=>n
la puterea a doua).

Answer 1

   
[tex]\displaystyle \\ n^2+n+41 = n(n+1) + 41 \\ n+1 = 41 \\ \Longrightarrow ~~~ n(n+1) + 41 = n \times 41 +41 = 41(n+1) = 41 \times 41=pp\\ \Longrightarrow ~~~ n = 41-1 = \boxed{40} \\ \\ Verificare: \\ n^2+n+41 = 40^2 + 40 + 41 =1600 + 40+41 = 1681 = 41^2 = pp[/tex]

[tex] sau \\ n^2+n+41 = 40^2 + 40 + 41 = 40(40+1) + 41 = \\ = 40 \times 41 + 41 = 41(40 +1) = 41 \times 41 = 41^2 = pp[/tex]