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

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If w is an even integer, then w^2 - 1 can be expressed as (w + 1)(w - 1), which shows it is the product of two consecutive integers. This means that for any even w greater than 2, w^2 - 1 cannot be prime since it will always be the product of two integers greater than 1. The only exception occurs when w equals 2, where w^2 - 1 equals 3, a prime number. The discussion suggests avoiding proof by contradiction and instead using the factorization approach to demonstrate the claim. Overall, the key takeaway is that w^2 - 1 is not prime for even integers except for the case when w equals 2.
dgamma3
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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.

prove by contradiction

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.
 
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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.
 
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