Question

Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60.
Find the numbers.

Answer 1

Let the three consecutive natural numbers be x, $x+1$ and $x+2$

According to the given condition,

$\therefore (x+1{)}^2-\left[\right(x+2{)}^2-x^2]=60$

$x^2+1+2x-\left[\right(x+2-x\left)\right(x+2+x\left)\right]=60$

$x^2+2x+1-\left[2\right(2+2x\left)\right]=60$

$x^2+2x+1-4-4x=60$

$x^2-2x-63=0$

$x^2-9x+7x-63=0$

$x(x-9)+7(x-9)=0$

$(x+7)(x-9)=0$

$\therefore x=9$ or -7

$\therefore x=9$

(neglect $x=-7$)

$\therefore$ Numbers are 9, 10, 11.