Question

va rog repede dau coroana​

Answer 1

$\displaystyle\bf\\m+n+p+m^2+n^2=28 \Leftrightarrow m(m+1)+n(n+1)+p=28,\\~m(m+1),~n(n+1)~sunt~pare.\\a)~m(m+1)+n(n+1)=28-p \Leftrightarrow 28-p=par \implies p=par,~dar~p~este~\\prim,~si~cum~singurul~numar~prim~par~este~2~rezulta~ca~p~este~2.\\b)~unul~dintre~doua~numere~naturale~consecutive~x~si~x+1~este~par.\\n~si~n+1~sunt~prime~daca~n=2,~daca~nu~n~sau~n+1~vor~fi~numere~pare~\\mai~mari~decat~2.\\m(m+1)+p=28-6 \Leftrightarrow m(m+1) +p =22,~iei~prin~incercari~m~de~la~1~\\pana~la~4.$

$\displaystyle\bf\\c)~m=n=2\implies p=16=4^2=patrat~perfect.$