# [{(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?

Jun 26, 2016

This proposition is false.

#### Explanation:

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

Then:

$\left[\left\{{\left(a + 2\right)}^{n} - {\left(a + 1\right)}^{n}\right\} - \left\{{\left(a + 1\right)}^{n} - \left({a}^{n}\right)\right\} \pm 1\right]$

$= \left[\left\{\left(a + 2\right) - \left(a + 1\right)\right\} - \left\{\left(a + 1\right) - \left(a\right)\right\} \pm 1\right]$

$= \left[1 - 1 \pm 1\right]$

$= \pm 1$

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

So the proposition is false.

Jun 26, 2016

The assertion is false

#### Explanation:

${\left(a + 2\right)}^{2 k + 1} + {a}^{2 k + 1} \pm 1 = a \cdot {p}_{2 k} \left(a\right) + {2}^{2 k + 1} \pm 1$
where ${p}_{2 k} \left(a\right)$ is an $2 k$ degree polynomial.

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

${\left(a + 2\right)}^{2 k + 1} + {a}^{2 k + 1} + 1 = a \cdot {p}_{2 k} \left(a\right) + m \cdot a$

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