Find 3 consecutive numbers where the product of the smaller two numbers is 19 less than the square of the largest number?

Let the numbers be ;- n, n+1, n+2.

Hence n(n+1) is 19 less than (n+2)^(2)

It follows;-

n(n+1)+ 19 = (n+2)^(2)

Multiply out the brackets.

n^(2) + n + 13 = n^(2) + 4n + 4

Collect 'like' terms.

n^(2) - n&(2) + 4n - n = 13 - 4

Add

3n = 9

n = 3

n+ 1 = 4

n +2 = 5

Hence 3,4, & 5 are the consecutive numbers.