Question 626301
find 3 consecutive positive odd integers such that the sum of the squares of the first and second integers is equal to the square of the third integer plus 33
<pre>
Smallest odd integer = x-2
Middle-sized odd integer = x
Largest odd integer = x+2

               (x-2)² + x² = (x+2)² + 33

      (x-2)² - (x+2)² = 33 - x²

[(x-2)-(x+2)][(x-2)+(x+2)] = 33-x²

        [x-2-x-2][x-2+x+2] = 33-x²

                  [-4][2x] = 33-x²

                       -8x = 33-x²

                  x²-8x-33 = 0

               (x-11)(x+3) = 0

                  x-11 = 0    x+3 = 0
                     x = 11     x = -3

We only want the positive odd integers so we 
ignore the negative answer.

Smallest odd integer = x-2 = 11-2 = 9
Middle-sized odd integer = x = 11
Largest odd integer = x+2 = 11+2 = 13

Edwin</pre>