Question

`[{(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?

Answer 1

This proposition is false.

Explanation:

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

The assertion is false

Explanation:

`(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.