Question

Aratati ca nu exista numere naturale n si p astfel incat sa fie devarata relatia n^2
- 3^p=2021

Answer 1

$\displaystyle\it\\un~patrat~perfect~poate~avea~una~din~formele~:~\mathcal{M}_3~sau~\mathcal{M}_3+1.\\Demonstratia~se~face~relativ~usor,~avand~la~cunostinta~ca~orice~\\numar~natural~poate~fi~de~forma:\mathcal{M}_3,~\mathcal{M}_3+1~sau~\mathcal{M}_3+2,~analizam~fiecare\\caz~si~ajungem~la~concluzia~de~mai~sus.\\------------------\\n^2-3^p=2021~|+3^p \implies n^2=2021+3^p.\\2021=\mathcal{M}_3+2 \implies daca~p\geq 1~atunci~2021+3^p=\mathcal{M}_3+2,~care\\nu~este~patrat~pefect,~deci~p~poate~fi~doar~0.$

$\displaystyle\it\\daca~p=0 \implies n^2=2022=\mathcl{M}_{10}+2,~care~nu~este~patrat~perfect.\\prin~urmare,~nu~exista~n,p\in\mathbb{N}~sa~verifice~relatia~de~mai~sus.$