# Is this a proof by cases?

• Oct 9th 2007, 04:31 AM
Is this a proof by cases?
Recall that a positive integer p is called prime if p is not equal to 1 and if the only positive integers which divide p are 1 and p itself. Prove that for all positive integers n, if n^3 +1 is a prime number, then n =1.

This seems to be a simple proof by cases. By substituting in 2m for even and then 2m + 1 for odd we get. (2m)^3 + 1= 8m^3 +1. Now this is where I get stuck. I think that by some sort of rule this must always divide something. Am I on the right track?

For the odd case we let n = 2m+1. As a result we get (2m+1)^3 + 1 which will always be even because when you multiply out inside the parenthesis you get an even + 1 then you add the 1 outside the parenthesis giving you an even number. This will as a result always divide 2 and because it is greater than 1 satisfies our second case.

I am probably missing something here. Any help would be greatly appreciated.
• Oct 9th 2007, 05:24 AM
topsquark
Quote:

Originally Posted by padsinseven
Recall that a positive integer p is called prime if p is not equal to 1 and if the only positive integers which divide p are 1 and p itself. Prove that for all positive integers n, if n^3 +1 is a prime number, then n =1.

This seems to be a simple proof by cases. By substituting in 2m for even and then 2m + 1 for odd we get. (2m)^3 + 1= 8m^3 +1. Now this is where I get stuck. I think that by some sort of rule this must always divide something. Am I on the right track?

For the odd case we let n = 2m+1. As a result we get (2m+1)^3 + 1 which will always be even because when you multiply out inside the parenthesis you get an even + 1 then you add the 1 outside the parenthesis giving you an even number. This will as a result always divide 2 and because it is greater than 1 satisfies our second case.

I am probably missing something here. Any help would be greatly appreciated.

It is much easier than this. Note that
\$\displaystyle n^3 + 1 = (n + 1)(n^2 - n + 1)\$

So for n = 1 we have \$\displaystyle 1^3 + 1 = 2\$ which is prime. Fo any other n the only possible factor that could equal 1 is the \$\displaystyle n^2 -n + 1\$ which is greater than 1 for n > 1. (Prove this.) Alternately you could show that for no other n than n = 1 is \$\displaystyle n + 1 = n^3 + 1\$ (and certainly n + 1 is greater than 1 for all n > 1.)

-Dan