SOLUTION: Three consecutive odd integers are such that the square of the third integer is 65 less than the sum of the squares of the first two. One solution is -7, -5, and -3. Find three oth

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Question 897396: Three consecutive odd integers are such that the square of the third integer is 65 less than the sum of the squares of the first two. One solution is -7, -5, and -3. Find three other consecutive odd integers that also satisfy the given conditions.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let N-2,N, and N%2B2 be the integers.
.
.
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%28N%2B2%29%5E2=N%5E2%2B%28N-2%29%5E2-65
N%5E2%2B4N%2B4=N%5E2%2BN%5E2-4N%2B4-65
N%5E2-8N-65=0
%28N-13%29%28N%2B5%29=0
Two solutions:
N-13=0
N=13
The integers are 11,13,and 15.
The other solution is already given.