[tex]1.~Restul~impartirii~unui~patrat~perfect~la~3~este~0~sau~1. \\ \\ Demonstratie:~x~poate~avea~una~din~formele: \\ \\ x=M_3 \Rightarrow x^2=M_3 \Rightarrow restul=0. \\ \\ x=M_3+1 \Rightarrow x^2=M_3+1^2=M_3+1 \Rightarrow restul=1. \\ \\ x=M_3+2 \Rightarrow x^2=M_3+2^2=M_3+4=M_3+1 \Rightarrow restul=1. \\ \\ Cum~numerele~nu~sunt~divizibile~cu~3,~rezulta~ca~restul~impartirii \\ \\ patratelor~lor~la~3~este~1. [/tex]
[tex]Deci~patratele~numerelor~sunt~de~forma~M_3+1. \\ \\ S=(M_3+1)+(M_3+1)+(M_3+1)+...+(M_3+1)= \\ \\ =M_3+2013= M_3+ 3 \cdot 671 =M_3+M_3=M_3. \\ \\ Notatia~"M_3"~semnifica~un~multiplu~natural~de-al~lui~3. \\ \\ Am~folosit~proprietatea~(a+b)^n=M_a+b^n.[/tex]
[tex]2.~Resturile~impartirii~unui~patrat~perfect~la~4~sunt~0~si~1. \\ \\ Demonstratie:~Orice~numar~natural~este~de~forma: \\ \\ x=2k \Rightarrowx^2=4k^2 \Rightarrow restul =0. \\ \\ x=2k+1 \Rightarrow x^2=4k^2+4k+1 \Rightarrow restul=1. \\ \\ Obsevam~ca~p=2~nu~indeplineste~cerinta. \\ \\ Analizam~cazul~p\ \textgreater \ 2,~iar~cum~p~este~prim,~rezulta~ca~p^2~nu~este~ \\ \\ divizibil~cu~4,~si,~prin~urmare,~restul~impartirii~lui~p^2~la~4~este~1. [/tex]
[tex]Deci~p^2=M_4+1.~Numarul~2^p~este~divizibil~cu~4~(pentru~ca~p\ \textgreater \ 2), \\ \\ deci~2^p=M_4. \\ \\ Prin~urmare~p^2+2^p=M_4+1+M_4=M_4+1. \\ \\ Insa~2015=4 \cdot 503+3=M_4+3. \\ \\ In~concluzie~ecuatia~nu~are~solutii. \\ \\ \underline{Observatie}:~ai~scris~gresit~enuntul.~Nu~era~"determinati",~ci \\ \\ "demonstrati".[/tex]
