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Question 692935: suppose you want to find two consecutive integers such that the sum of their squares is 613. what are they?
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! Let the two integers be n and n+1
n^2 + (n+1)^2 = 613
n^2 + n^2 + 2n + 1 = 613
2n^2 + 2n - 612 = 0
n^2 + n - 306 = 0
This can be factored as:
(n-17)(n+18) = 0
This gives n = 17, n = -18
So there are two possible solutions: 17, 18 and -18, -17
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