# If w is an even integer, then w^2 - 1 is not a prime number.

1. Aug 31, 2012

### dgamma3

hello, I am trying to solve this problem:
If w is an even integer, then w^2 - 1 is not a prime number.

my current working.

If w is a even integer then w^2 -1 is a prime number.

if w = 2x
then $w^{2}$ -1
= $4x^{2}$ -1

I am not sure where to go from here, maybe congruence relations:

n+1 = ($4x^{2}$)y
($4x^{2}$)y | n+1 therefore this is a contradiction.

is this correct.
thanks.

2. Aug 31, 2012

### conquest

Don't do it by contradiction, but use:

w²-1=(w+1)(w-1)

also there is the exception of w=2 since then (w+1)(w-1)=3 $\cdot$ 1=3 which is of course a prime. But by splitting it up this way we already see why. either side is the product of primes but this time one side is the empty product. But this is only true if w=2 since the integers as a ring have but one unit for multiplication.