Question 256445
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Let *[tex \Large x] represent the smaller of the two consecutive odd integers.  Then *[tex \Large x\,+\,2] must represent the next consecutive odd integer.


The reciprocal of the smaller is then *[tex \Large \frac{1}{x}], and the reciprocal of the larger one is *[tex \Large \frac{1}{x\,+\,2}].  Then the difference between the reciprocals is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x}\ -\ \frac{1}{x\,+\,2}\ =\ \frac{2}{63}]


Combine the fractions in the LHS using the common denominator *[tex \Large x^2\,+\,2x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(x\,+\2\right)\ -\ x}{x^2\,+\,2x}\ =\ \frac{2}{63}]


Collect terms on the left:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{x^2\,+\,2x}\ =\ \frac{2}{63}]


Now we have two equal fractions with equal numerators -- therefore the denominators must be equal as well, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\,+\,2x\ =\ 63]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\,+\,2x\,-\,63\ =\ 0]


Solve the factorable quadratic.  Discard the negative root.  The positive root will be the smaller of the two positive consecutive odd integers.


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