#[{(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?

2 Answers
Jun 26, 2016

Answer:

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.

Jun 26, 2016

Answer:

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.