Question 245451
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Let *[tex \Large x] represent the first integer.


Then *[tex \Large x\ +\ 1] is the next consecutive integer.  But if *[tex \Large x] is even, then *[tex \Large x\ +\ 1], must be odd.  Therefore, the next consecutive <b><i>even</i></b> integer must be *[tex \Large x\ +\ 2].


It follows then that the next consecutive integer after that must be *[tex \Large x\ +\ 4]


Square the first integer:  *[tex \Large x^2]


Square the second integer:  *[tex \Large (x\ +\ 1)^2\ =\ x^2\ +\ 4x\ + 4]


Add these two quantities: *[tex \Large 2x^2\ +\ 4x\ + 4]


And this sum is equal to the square of the third integer:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 4x\ + 4\ =\ x^2\ +\ 8x\ +\ 16]


Collect like terms in the LHS, leaving the RHS equal to zero:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 4x\ -\ 12\ =\ 0]


Now solve the factorable quadratic.  Each of the roots will be the first of a series of three consecutive even integers that fit the parameters of the problem.  Remember that 0 is a perfectly good even integer in this context.


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