Question 353429
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The sum of the squares of three consecutive positive integers is 77. Find the integers.
The book gives a hint: if one integer is x, the next consecutive positive integer is x+1, and the third is x+2
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        There is much more elegant way to solve, than the way assumed in the book.



<pre>
Let 'n' be the central number of these three consecutive integers.

So, our numbers are (n-1), n and (n+1).


Make an equation for the sum of squares

    (n-1)^1 + n^2 + (n+1)^2 = 77,

    (n^2 - 2n + 1) + n^2 + (n^2 + 2n + 1) = 77,


Cancel opposite terms '-2n' and '2n' and combine like terms

    3n^2 + 2 = 77  --->  3n^2 = 77 - 2  --->  3n^2 = 75  --->  n^2 = 75/3 = 25  --->  n = {{{sqrt(25)}}} = +/- 5.


Since we are looking for positive integer numbers, they are 4, 5 and 6.    <U>ANSWER</U>
</pre>

Solved, without necessity to solve a quadratic equation.