Question 1037586
Three consecutive odd integers are such that the square of the third integer is 153 less than the sum of the squares of the first two. One solution is -11,-9, and -7. Find three other consecutive odd integers that also satisfy the given condition. 
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(n-2)^2 + n^2 = (n+2)^2 + 153
2n^2 - 4n + 4 = n^2 + 4n + 157
n^2 - 8n - 153 = 0
(n - 17)*(n + 9) = 0
--> 15, 17 & 19