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

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SUMMARY

The discussion centers on proving that if w is an even integer, then w² - 1 is not a prime number. The initial approach suggested proving by contradiction, assuming w² - 1 is prime for even w. However, a more effective method involves factoring w² - 1 into (w + 1)(w - 1). The exception of w = 2 is noted, where w² - 1 equals 3, a prime number. This analysis confirms that for all even integers greater than 2, w² - 1 cannot be prime due to its factorization.

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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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