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Three consecutive odd integers are such that the square of the third integer is seven greater than the sum of the squares of the first two. One solution is -1, 1, and 3. Find three other consecutive odd integers that also satisfy the given conditions

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Let the first odd integer be x.  Then the next odd integer will be x+2, and the next x+4.

"The sq. of the 3rd is 7 greater than the sum of the squares of the first two" translates into:

         (x+4)^2 = 7 + (x)^2 + (x+2)^2

Expanding,    x^2 + 8x + 16 = 7 + x^2 + x^2 + 4x + 4

Cross out x^2 on both sides:   8x + 16 = 7 + x^2 + 4x + 4

Subtract 16 from both sides:     8x         = 7 + x^2 + 4x  + 4 -16
       
                                             or:     8x         = x^2 + 4x + 11 - 16

                                                       8x        = x^2 + 4x - 5
       
Subtracting 8x from both sides:      0         = x^2 -4x - 5

Solve this by completing the square:

                                                         0 = x^2 - 4x + 4  -4 - 5, or

                                                          0 = (x-2)^2 - 9

                                 Thus,               x-2 = plus or minus sqrt(9)

                                                         x-2 = plus or minus 3

                                                         x = 2 plus or minus 3
 
                                                          x = 5 or x = -1

Try x = 5.  Then the consec. odd integers are {5, 7, 9} (answer)

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