Question 187552
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Let the first integer be <i>x</i>, then the second is <i>x</i> + 1, the third is <i>x</i> + 2, and so on.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 + (x + 1)^2 + (x + 2)^2 = (x + 3)^2 + (x + 4)^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = (x^2 + 6x + 9) + (x^2 + 8x + 16)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2 + 6x + 5 = 2x^2 + 14x + 25]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 8x -20 = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - 10)(x + 2) = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 10 = 0 \ \ \Rightarrow\ \ x = 10]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x + 2 = 0 \ \ \Rightarrow\ \ x = -2]


Exclude the negative value because we are looking for positive integers.  So the first integer is 10 and the others are therefore 11, 12, 13, 14.


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10^2 + 11^2 + 12^2 =^? 13^2 + 14^2]


You can do your own arithmetic to verify.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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