SOLUTION: Find three consecutive positive integers which have 110 as the sum of their squares.

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Question 1197045: Find three consecutive positive integers which have 110 as the sum of their squares.
Answer by ikleyn(52848) About Me  (Show Source):
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Find three consecutive positive integers which have 110 as the sum of their squares.
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(x-1)^2 + x^2 + (x+1)^2 = 110


(x^2 - 2x + 1) + x^2 + (x^2 + 2x + 1) = 110


3x^2 + 2 = 110


   3x^2  = 110 - 2 = 108


    x^2            = 108/3 = 36


    x                      = sqrt%2836%29 = 6.


ANSWER.  The numbers are 5, 6 and 7.


CHECK.  25 + 36 + 49 = 110.    ! correct !

Solved.