√n²-2n+16∈N ⇔n²-2n+16=y² unde y∈N;
n²-2n+1+15=y² ⇔(n-1)²+15=y² ⇔(y-n+1)(y+n-1)=15
Consideram urmatoarele variante;
a. y-n+1=1 ⇒y-n=0
y+n-1=15 ⇒y+n=16⇒y=n=8;
b. y-n+1=3 ⇒y-n=2
y+n-1=5 ⇒y+n=6⇒y=4 si n=2;
c. y-n+1=5⇒y-n=4
y+n-1=3⇒y+n=4 ⇒ceea ce este imposibil;
d. y-n+1=15 ⇒y-n=14
y+n-1=1⇒y+n=2 ⇒ceea ce este imposibil;
Asadar n∈{2;8}.
