Vom arata ca in general n(n+1)(n+2)(n+3)+1 este patrat perfect.
n(n+1)(n+2)(n+3)+1=n(n+3)x(n+1)(n+2)+1=(n²+3n)(n²+3n+2)+1=
=(n²+3n)(n²+3n)+2(n²+3n)+1²=(n²+3n)²+2(n²+3n)+1²
=(n²+3n+1)²
n(n+1)(n+2)(n+3)+1=(n²+3n+1)²
Pentru n=2012 obtinem:
2012*2013*2014*2015+1=(2012²+3·2012+1)²
