SOLUTION: Three consecutive odd integers. The sum of the square of the first and the square of the third is 8 more than twice the square of the second.

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Question 284320: Three consecutive odd integers.
The sum of the square of the first and the square of the third is 8 more than twice the square of the second.
Answer by dabanfield(803) About Me  (Show Source):
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Three consecutive odd integers.
The sum of the square of the first and the square of the third is 8 more than twice the square of the second.
Let x be the smallest of the three integers. Then the next two odd integers are x+2 and x+4.
So we have:
x^2 + (x+4)^2 = 2*(x+2)^2 + 8
x^2 + (x^2 + 8x + 16) = 2*(x^2+4x+4) + 8
x^2 + x^2 + 8x + 16 = 2x^2 + 8x + 8 + 8
2x^2 + 8x + 16 = 2x^2 + 8x + 16
Not that the above is true for all x and consequently for any three consecutive odd integers.