Question 732899
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The product of two even numbers is even.  The product of two odd numbers is odd.  The sum of two even numbers is even and the sum of two odd numbers is also even.


Since the product of two even numbers is even, the cube of an even number must be even.  Then the sum of the even number cubed (even) plus the even number itself must be even.


Since the product of two odd numbers is odd, the cube of an odd number must be odd. Then the sum of the odd number cubed (odd) plus the odd number itself must be even.


Since all positive integers are either even or odd, *[tex \LARGE 2\ \mid\ n^3\ +\ n]. Q.E.D.


Indeed, by extension:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \forall\,n,\,a,\,b\,\in\,\mathbb{Z}\ :\ n\,>\,0,\ a\,>\,0,\ b\,>\,0] it can be said that *[tex \LARGE 2\,\mid\,n^a\ +\ n^b]


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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