[tex]n^{2}+2018=p^{2},~n,p \in \mathb{N}~?[/tex]
[tex]Presupunem~ca~exista~numerele~naturale~n,~p~astfel~incat~n^{2}+2018=p^{2}[/tex]
[tex] \Rightarrow p^{2}-n^{2}=2018[/tex]
[tex] \Rightarrow (p-n)(p+n)=2018[/tex]
[tex]2018~scris~ca~produs~de~doua~numere~naturale:~2018=2\cdot 1009=1\cdot 2018[/tex]
[tex]\Rightarrow \left \{ {{p-n=2} \atop {p+n=1009}} \right. ~sau:~ \left \{ {{p-n=1009} \atop {p+n=2}} \right. ~sau:~ \left \{ {{p-n=1} \atop {p+n=2018}} \right. ~sau:~ \left \{ {{p-n=2018} \atop {p+n=1}} \right. [/tex]
[tex]Caz~I:~p-n=2 \Rightarrow p=n+2\\
\\ p+n=1009\\
\\ n+2+n=1009\\
\\ 2n=1007\\
\\ n=1003,5~nu~apartine~numerelor~naturale[/tex]
[tex]Caz~II:~p-n=1009 \Rightarrow p=n+1009\\
\\ p+n=2\\
\\ n+1009+n=2\\
\\ 2n=-1007\\
\\ n=-1003,5~nu~apartine~numerelor~naturale[/tex]
[tex]Caz~III:~p-n=1 \Rightarrow p=n+1\\
\\ p+n=2018\\
\\ n+1+n=2018\\
\\ 2n=2017\\
\\ n=1008,5~nu~apartine~numerelor~naturale[/tex]
[tex]Caz~IV:~p-n=2018 \Rightarrow p=n+2018\\
\\ p+n=1\\
\\ n+2018+n=1\\
\\ 2n=-2017\\
\\ n=-1008,5~nu~apartine~numerelor~naturale[/tex]
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