Thread: 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?

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).

undefined has shown that it needs just one case.

Following on from your logic....

n even
n=2c



which is the product of two factors, hence non-prime.

Thanks guys. I was making it way more difficult than it had to be, as I so often do