SOLUTION: Three consecutive odd integers are such that the square of the third integer is 105 less than the sum of the squares of the first two. One solution is -9, -7, and -5. Find three

Question 353647: Three consecutive odd integers are such that the square of the third integer is 105 less than the sum of the squares of the first two. One solution is -9, -7, and -5. Find three other consecutive odd integers that only satisfy the given conditions.
Answer by sudhanshu_kmr(1152) (Show Source): You can put this solution on YOUR website!

Let integers are x, x+2, x+4

(x+4)^2 = x^2 + (x+2)^2 - 105
=> x^2 + 8x + 16 = x^2 + x^2 + 4x + 4 - 105
=> x^2 - 4x - 117 = 0
=> x^2 -13x + 9x -117 = 0
=> x(x-13) + 9( x-13) = 0
=> ( x+9) (x-13) = 0

x = -9 or x = 13
other consecutive integers are 13, 15, 17.

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