$[{(a+2)^n - (a+1)^n} - {(a+1)^n - (a^n)} pm 1]$ is a prime number when $a$ is a real number and $n$ is a positive odd number. How do we proof it?

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Answer 1 · 2016-06-26T19:24:35

This proposition is false.

Consider the case $n=1$ (which is an odd positive number).

Then:

$[{(a+2)^n-(a+1)^n}-{(a+1)^n-(a^n)}+-1]$

$=[{(a+2)-(a+1)}-{(a+1)-(a)}+-1]$

$=[1-1+-1]$

$=+-1$

Neither $1$ nor $-1$ are prime.

So the proposition is false.

Answer 2 · 2016-06-26T20:43:08

The assertion is false

$(a+2)^{2k+1}+a^{2k+1}pm1 = a cdot p_{2k}(a)+2^{2k+1}pm1$
where $p_{2k}(a)$ is an $2k$ degree polynomial.

But $2^{2k+1}+ 1$ is divisible by $3$ so for $a = 3$

$(a+2)^{2k+1}+a^{2k+1}+1 = a cdot p_{2k}(a)+m cdot a$

being divisible for $a$ . So the assertion is false.

[{(a+2)^n - (a+1)^n} - {(a+1)^n - (a^n)} pm 1] [{(a+2)^n - (a+1)^n} - {(a+1)^n - (a^n)} pm 1] is a prime number when aa is a real number and nn is a positive odd number. How do we proof it?