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 middle number of the three consecutive numbers is x, the other two numbers are $x-1$ and $x+1$.
According to the given condition, we have
x² = [(x +1)² -(x -1)²] +60
x² = (x² +2x +1) -(x² -2x +1) +60 = 4x +60
x² -4x -60 = 0

=> (x -10)(x +6) = 0
x = 10 or -6
Since x is a natural number, we get x = 10
Hence the three numbers are,

$x-1=10-1=9$,

$x=10$

$x+1=10+1=11$