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

  • Thread starter dgamma3
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  • #1
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Main Question or Discussion Point

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 [itex]w^{2}[/itex] -1
= [itex]4x^{2}[/itex] -1

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

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

is this correct.
thanks.
 

Answers and Replies

  • #2
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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 [itex]\cdot[/itex] 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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