Fie A=n(n+1)(n+2)(n+3); trebuie observat ca in cazul a oricaror patru numere naturale consecutive (n+1)(n+2)-n(n+3)=2;
- notam P=(n+1)(n+2)-1 ⇒ (n+1)(n+2)=P+1 ⇒ n(n+3)=P-1;
A=n(n+1)(n+2)(n+3)=(P+1)(P-1)=P²-1 ceea ce trebuia demonstrat! Cheia rezolvarii sta in observatia facuta!
Sau:
x(x+1)(x+2)(x+3)+1=[x(x+3)][(x+1)(x+2)]==(x²+3x)(x²+3x+2)+1=(x²+3x)[(x²+3x)+2]+1=
=(x²+3x)²+2(x²+3x)+1=[(x²+3x)+1]²=(x²+3x+1)²
