[tex]In~tipul~acesta~de~ probleme~ai~doua~tipuri~de~rezolvari: ~cea~pe~care\\ \\ai~ folosit-o,~si~inca~una...pe~care~o~voi~prezenta~la~final. \\ \\ Ai~pornit~bine,~insa~ai~finalizat~gresit. \\ \\ (a+2n+5)(a-2n-5)=40 \\ \\ Ideea~e~ca~ambii~factori~sunt~numere~intregi,~si~deci~sunt~divizori \\ \\ de-ai~lui~40.~Mai~mult:~a+2n+5\ \textgreater \ 0,~si,~deci, ~a-2n-5\ \textgreater \ 0.\\ \\ Si~mai~mult:(a+2n+5)~si~(a-2n-5)~au~aceeasi~paritate!~...~si~ \\ \\cum~40~este~par,~rezulta~ca~cei~doi~factori~sunt~pari. [/tex]
[tex]Mult~mai~mult:~a+2n+5\ \textgreater \ a-2n-5~(asta~pentru~ca~n \geq 0). \\ \\ Si~...~a+2n+5 \geq 5. \\ \\ Pe~baza~acestor~numeroase~observatii~pe~care~le~folosesc~de~cate \\ \\ori~am~ocazia,~pentru~a~evita~numeroasele~cazuri~de~verificat~(de~ \\ \\ exemplu:~in~aceasta~problema,~fara~aceste~observatii~ar~fi~fost \\ \\ 16~cazuri~de~analizat~(eventual~8...)).\\ \\ Revenind...~Avem~deci~urmatoarele~cazuri:[/tex]
[tex]i)~ \left \{ {{a+2n+5=10} \atop {a-2n-5=4}} \right. \Rightarrow~a=7~si~n=-1,~nu~convine! \\ \\ ii) \left \{ {{a+2n+5=20} \atop {a-2n-5=2}} \right. \Rightarrow ~ a=11~si~n=2.\\ \\ Deci~singura~solutie~este~ \boxed{n=2} .[/tex]
[tex]Cealalta~metoda~promisa:~ \\ \\ Avem:~a^2=4n^2+20n+65. ~Observam~ca~n=0~nu~este~solutie, \\ \\deci~vom~analiza~cazul~n \geq 1. \\ \\ Vom~incerca~sa-l~incadram~pe~4n^2+20n+65~intre~doua~patrate~ \\ \\perfecte~(nu~neaparat~consecutive). \\ \\ 4n^2+20n+65=(2n)^2+2 \cdot (2n) \cdot 5+5^2+40=(2n+5)^2+40. \\ \\ Deci~4n^2+20n+65\ \textgreater \ (2n+5)^2 \Leftrightarrow a^2\ \textgreater \ (2n+5)^2\\ \\(2n+6)^2=4n^2+24n+36,hmm... (nu~tocmai~util)\\ \\ (2n+7)^2=4n^2+28n+49 ~(din~nou...nu~tocmai~util).[/tex]
[tex](2n+8)^2=4n^2+32n+64\ \textgreater \ 4n^2+20n+65=a^2. \\ \\ Deci~(2n+8)^2\ \textgreater \ a^2\ \textgreater \ (2n+5)^2. \\ \\ Se~rezolva~pe~rand~ecuatiile~a^2=(2n+7)^2~si~a^2=(2n+6)^2,~\\ \\tinandu-se~cont~de~faptul~ca~a^2=4n^2+20n+65 .[/tex]
