Question

aratati ca exista n ne natural astfel incat n^2+n+41sa fie patrat perfect

Answer 1

$Fie~k \in N~astfel~incat~n^2+n+41=k^2. \\ \\ Avem:~4n^2+4n+164=4k^2\\ \\ (4n^2+4n+1)+163=(2k)^2 \\ \\ (2n+1)^2+163=(2k)^2 \\ \\163=(2k)^2-(2n+1)^2 \\ \\163=(2k+2n+1)(2k-2n-1)$

$Observam~ca~163~este~prim,~si~2k+2n+1\ \textgreater \ 2k-2n-1,~deci \\ \\ avem~ \left \{ {{2k+2n+1=163} \atop {2k-2n-1=1}} \right. .~Insumand,~obtinem~4k=164 \Rightarrow k=41 \Rightarrow \\ \\ \Rightarrow \boxed{n=40} \Rightarrow~exista~n \in N~astfel~incat~n^2+n+41.\\ \\ \underline{Observatii}\\ \\-nu~doar~ca~exista~un~asemenea~"n",~dar~el~este~si~unic!~(deoarece~ \\ \\ 163~este~prim)$

$-functia~g:N \rightarrow N,~g(n)=n^2+n+41~"genereaza"~doar~numere \\ \\ prime~pentru~orice~n \in~\{0,1,2,...,39 \}~(Euler)$