SOLUTION: Find three even consecutive integers such that the sum of the squares of the first and second integers is equal to the square of the third integer plus 48.

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Question 1029131: Find three even consecutive integers such that the sum of the squares of the first and second integers is equal to the square of the third integer plus 48.
Answer by mananth(16946) About Me  (Show Source):
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let three even consecutive integers be x , x+2,x+4
such that the sum of the squares of the first and second integers
x^2+(x+2)^2
is equal to the square of the third integer plus 48.
(x+4)^2
x^2+(x+2)^2=(x+4)^2+48
x^2+x^2+4x+4 =x^2+8x+16+48
x^2-4x-60=0
x^2-10x+6x-60=0
x(x-10)+6(x-10)=0
(x-10)(x+6)=0
x=-6 OR x=10
-6,-4,-2
OR
10,12,14