# Prove that n^2 - 1

• Sep 14th 2010, 04:09 PM
dblaisewatson
Prove that n^2 - 1
Hey everyone, I have this problem for homework and got started on it when n is odd (easy). But can't figure out how to solve it when n is even:

Let n be an integer greater than or equal to 3. Prove that n^2 -1 is not prime.

So I split it into when n is odd and when n even.

ODD:

n = 2c + 1
n^2 - 1 = (2c + 1)^2 - 1
"" = 4c^2 + 4c + 1 - 1
"" = 2(2c^2 + 2c)
d = 2c^2 + 2c
"" = 2d

Therefore 2 | n^2 -1, so n^2 -1 isn't prime for odd numbers

But I can't figure out where to go from here for even n's. Any hints on how to think about this?
• Sep 14th 2010, 04:19 PM
undefined
Quote:

Originally Posted by dblaisewatson
Hey everyone, I have this problem for homework and got started on it when n is odd (easy). But can't figure out how to solve it when n is even.

So I split it into when n is odd and when n even.

ODD:

n = 2c + 1
n^2 - 1 = (2c + 1)^2 - 1
"" = 4c^2 + 4c + 1 - 1
"" = 2(2c^2 + 2c)
d = 2c^2 + 2c
"" = 2d

Therefore 2 | n^2 -1, so n^2 -1 isn't prime for odd numbers

But I can't figure out where to go from here for even n's. Any hints on how to think about this?

You can handle even and odd at once by factoring n^2 - 1 (think: difference of squares).
• Sep 14th 2010, 04:29 PM
undefined has shown that it needs just one case.

$n^2-1=4c^2-1=(2c+1)(2c-1)$