Question 63910
Find two consecutive positive numbers such that the product of the sum and difference of the numbers plus eight is the sum of their squares.
<p>
Let n = the first of the two numbers.
Then n+1 = the second of the numbers.

The equation to solve is {{{(n+(n+1))(n-(n+1))+8 = n^2+(n+1)^2}}}.
This expands to {{{(2n+1)(-1) + 8 = 2n^2+2n+1}}}.
Or, {{{-2n+7 = 2n^2+2n+1}}}.
Or, {{{7 = 2n^2+4n+1}}}.
Or, {{{6 = 2n^2+4n}}}.
Or, {{{3 = n^2+2n}}}.
Or, {{{n^2+2n-3 = 0}}}.

This factors: {{{(n+3)(n-1) = 0}}}.
So, n=-3 or n=1. 
But, the problem states that the numbers are positive.
So, n=1. The second number is n+1 = 2.

We verify this answer by noting that the sum of the numbers (1+2=3) times the difference of the numbers (1-2=-1) plus 8 = 3(-1)+8 = 5 which is indeed the sum of 1 squared plus 2 squared.