Question 1014564
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Let *[tex \Large x] represent the smaller of the two odd integers, then *[tex \Large x\ +\ 2] must be the next consecutive odd integer.


The product of the two consecutive odd integers is *[tex \Large x(x\ +\ 2)\ =\ x^2\ +\ 2x]


Two times the sum of the two consecutive odd integers is *[tex \Large 2(x\ +\ x\ +\ 2)\ =\ 4x\ +\ 4]


So, if the product is 31 larger than twice the sum:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ -\ 31\ =\ 4x\ +\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 2x\ -\ 35\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 7)(x\ +\ 5)\ =\ 0]


So if *[tex \Large x\ =\ 7], then *[tex \Large x\ +\ 2\ =\ 9]


7 times 9 is 63 which is 31 larger than 32 which is 2 times 16 which is 7 plus 9


But if *[tex \Large x\ =\ -5], then *[tex \Large x\ +\ 2\ =\ -3]


-5 times -3 is 15 which is 31 larger than -16 which is 2 times -8 which is -5 plus -3.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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