Question 245504
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Let *[tex \Large x] represent the first integer.  Then the next consecutive integer is *[tex \Large x\ +\ 1].


The square of the second:  *[tex \Large (x\ +\ 1)^2\ =\ x^2\ + 2x\ +\ 1]


6 times the first: *[tex \Large 6x]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ + 8x\ +\ 1\ =\ 181]


Put in standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ + 8x\ -\ 180\ =\ 0]


Solve the factorable quadratic.  Exclude the negative root because the problem asks for a positive integer.  The positive root is the first integer, one more than that is the second one.


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