Question 682679
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Let *[tex \LARGE x] represent any positive odd integer.  Then *[tex \LARGE x\ +\ 2] is the next consecutive positive odd integer.


The sum of these two positive odd integers is *[tex \LARGE 2x\ +\ 2]


Assume that *[tex \LARGE 2x\ +\ 2] is prime.  But *[tex \LARGE 2x\ +\ 2] is divisible by 2 and therefore even.  The only even prime number is 2, hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ 2\ =\ 2]


Which means that *[tex \LARGE x\ =\ 0].  But *[tex \LARGE 0] is neither positive nor odd, contradicting the original assumption about the nature of *[tex \LARGE x].  Therefore, reductio ad absurdum, *[tex \LARGE 2x\ +\ 2] cannot be prime for any positive odd value of *[tex \LARGE x].  Q.E.D.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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