[tex]\text{Tipul acesta de exercitii se pot rezolva in doua moduri:}\\
Metoda\ 1:
\text{Deoarece }n^2+4n-5 \text{ este patrat perfect,}\exists\ p\in \mathbb{Z}\ \text{astfel}\\ \text{incat } n^2+4n-5=p^2\\
n^2+4n-5-p^2=0\\
\Delta=16+20+4p^2=36+4p^2\\
\text{Solutii ecuatiei sunt numere intregi,}\text{deci }\sqrt{\Delta}\in \mathbb{N} \\
\text{Din nou,} \exists\ m\in \mathbb{Z}\ \text{astfel incat}\ \Delta=m^2\\
36+4p^2=m^2\\
m^2-4p^2=36\\
(m-2p)(m+2p)=36\\
\text{Mai departe cautam toate variantele posibile,tinand cont ca:}\\ m-2p \text{ si }m+2p \text{au aceeasi paritate}[/tex]
[tex] \left \{ {{m-2p=2} \atop {m+2p=18}} \right. \\
-4p=-16\Rightarrow p=4\\
\left \{ {{m-2p=18} \atop {m+2p=2}} \right\\
-4p=16\Rightarrow p=-4\\
\text{In fine,mai ai de analizat cazurile (-2,-18),(-18,-2),(6,6),(-6,-6)}\\
\text{Il inlocuiesti pe p si il afli pe n}.\\
\\
Metoda\ 2\\
\text{Mergand pe aceeasi idee ca la prima metoda,presupunem ca }\exists p\in \mathbb{Z}\\ \text{astfel incat }n^2+4n-5=p^2\\
\text{Facem un artificiu de calcul:}\\
(n+2)^2-9=p^2\\
(n+2)^2-p^2=9\\
(n+2-p)(n+2+p)=9[/tex]
[tex]\text{Din nou avem de analizat mai multe variante:}\\
\left \{ {{n+2-p=1} \atop {n+2+p=9}} \right.\\
2n+4=10\Rightarrow n=3\\
\left \{ {{n+2-p=3} \atop {n+2+p=3}} \right. \\
2n+4=6\Rightarrow n=1\\
\left \{ {{n+2-p=-3} \atop {n+2+p=-3}} \right.\\
2n+4=-6\Rightarrow n=-5\\
\left \{ {{n+2-p=-9} \atop {n+2+p=-1}} \right. \\
2n+4=-10\Rightarrpw n=-7\\
\text{Si am obtinut toate solutiile.}[/tex]
