[tex]\dispalystyle\it\\d)~\sqrt{n^2-n-11} \in\mathbb{N} \Leftrightarrow \exists k\in\mathbb{N}~a.i.~n^2-n-11=k^2.\\\\n^2-n-11=k^2 |\cdot4 \implies 4n^2-4n-44=4k^2 \Leftrightarrow\\\\4n^2-4n+1-45 =4k^2 \Leftrightarrow (2n-1)^2-45=4k^2 \Leftrightarrow\\\\(2n-1)^2-(2k)^2=45 \Leftrightarrow (2n-1-2k)(2n-1+2k)=45,casework.\\45=(-1)(-45)=(-45)(-1)=45\cdot1=1\cdot45=(-15)(-3)=\\(-3)(-15)=15\cdot3=3\cdot15=(-5)(-9)=(-9)(-5)=\\5\cdot9=9\cdot5.\\\\pentru~ca~sunt~prea~multe~cazuri,~incercam~sa~mai~reducem\\din~ele.[/tex]
[tex]\displaystyle\it\\daca~(n,k)~este~solutie~atunci~si~(n,-k)~este~solutie,~\\atunci~putem~prespune~fara~a~restrange~generalitatea~\\ca~k\geq 0 \implies 2n-1+2k\geq 2n-1-2k.\\\\deci,~raman~cazurile:45=(-45)(-1)=(-15)(-3)=(-9)(-5)=\\1\cdot45=3\cdot15=5\cdot9.[/tex]
