Question 1154151
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Find three consecutive positive odd integers such that the sum of their squares is 371.
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        I will show another way to calculate.

        It requires less calculations and is much more educative than shown by the other person.



<pre>
Let n be the central number, so the numbers are (n-2), n and (n+2).


Our equation is

    (n-2)^2 + n^2 + (n+2)^2 = 371.


Simplify and find 'n'

    (n^2 - 4n + 4) + n^2 + (n^2 + 4n + 4) = 371

     3n^2 + 8 = 371

     3n^2 = 371 - 8 = 363

      n^2 = 363/3 = 121

      n = {{{sqrt(121)}}} = +/- 11.


We are looking for positive integers, so n = 11,  and the numbers are 9, 11, 13.
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Solved.


It is how this problem is EXPECTED to be solved.