Question

DAU COROANA
EXISTA NUMERELE NATURALE n SI p ASTFEL INCAT
n la 2 +2018=p la 2
dau 35 punte total

Answer 1

$n^{2}+2018=p^{2},~n,p \in \mathb{N}~?$

$Presupunem~ca~exista~numerele~naturale~n,~p~astfel~incat~n^{2}+2018=p^{2}$

$\Rightarrow p^{2}-n^{2}=2018$

$\Rightarrow (p-n)(p+n)=2018$

$2018~scris~ca~produs~de~doua~numere~naturale:~2018=2\cdot 1009=1\cdot 2018$

$\Rightarrow \left \{ {{p-n=2} \atop {p+n=1009}} \right. ~sau:~ \left \{ {{p-n=1009} \atop {p+n=2}} \right. ~sau:~ \left \{ {{p-n=1} \atop {p+n=2018}} \right. ~sau:~ \left \{ {{p-n=2018} \atop {p+n=1}} \right.$

$Caz~I:~p-n=2 \Rightarrow p=n+2\\ \\ p+n=1009\\ \\ n+2+n=1009\\ \\ 2n=1007\\ \\ n=1003,5~nu~apartine~numerelor~naturale$

$Caz~II:~p-n=1009 \Rightarrow p=n+1009\\ \\ p+n=2\\ \\ n+1009+n=2\\ \\ 2n=-1007\\ \\ n=-1003,5~nu~apartine~numerelor~naturale$

$Caz~III:~p-n=1 \Rightarrow p=n+1\\ \\ p+n=2018\\ \\ n+1+n=2018\\ \\ 2n=2017\\ \\ n=1008,5~nu~apartine~numerelor~naturale$

$Caz~IV:~p-n=2018 \Rightarrow p=n+2018\\ \\ p+n=1\\ \\ n+2018+n=1\\ \\ 2n=-2017\\ \\ n=-1008,5~nu~apartine~numerelor~naturale$

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