Answer to Question #89794 in Algebra for Aravind

Question

For any real number x, prove that x(x+1)(x+2)(x+3)+1 ≥ 0

Answer 1

x(x+1)(x+2)(x+3)+1=(x+1)(x+2)[x(x+3)]+1=(x²+2x+x+2)[x(x+3)]+1=

(x²+3x+2)[x(x+3)]+1=[x(x+3)+2][x(x+3)]+1=

[x(x+3)]²+2[x(x+3)]+1=[x(x+3)+1]².

So

x(x+1)(x+2)(x+3)+1 = [x(x+3)+1]².

Consequently x(x+1)(x+2)(x+3)+1 is a perfect square.

As x(x+1)(x+2)(x+3)+1 is a perfect square, a perfect square for real number

is always non-negative ( ≥ 0).