Question

If the product of four consecutive natural numbers increased by a natural number p is a perfect square, then the value of p is

• A. 8
• B. 4
• C. 2
• D. 1

Answer 1

The correct option is D 1

Let the four consecutive natural numbers be $x,x+1,x+2$ and $x+3$. Then,
A perfect square $=x(x+1)(x+2)(x+3)+p$
$=x(x+3)(x+1)(x+2)+p$
$=(x^2+3x)\times (x^2+3x+2)+p$
$=(x^2+3x{)}^2+2(x^2+3x+2)+p$
$=(x^2+3x{)}^2+2(x^2+3x)+p$
The expression on the right hand side will be a perfect square if and only p =1
Perfect square number
$=\left[\right(x^2+3x{)}^2+2(x^2+3x)+1]$
$=(x^2+3x+1{)}^2$