- #1
canningdevin
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If p >= 5 is prime, prove that p^2 + 2 is composite.
So i noticed if we divide any p >= 5 by 6 we only get remainders of 1 or 5.
6 | 5 , r = 5
6 | 7 , r = 1
6 | 11, r = 5
6 | 13, r = 1
6 | 17, r = 5 and so on
so for my proof i am saying for p >= 5, p = 6k + 1 or 6k = 5
so for the first , p^2 + 2 = (6k+1)^2 + 2 = 36k^2 +12k + 3 = 3(12k^2 + 4k + 1)
and similarly, p^2 + 2 = (6k+5)^2 + 2 = 36k^2 + 60k + 27 = 3(12k^2 + 20k + 3)
so both of these are composite so that is great.
But is there a way i can justify that any prime p >= 5 can be written as 6k + 1 or 6k + 5?
I mean i can go through a ton of examples and the remainders are only 1 or 5 but examples don't mean anything.
Can someone help me justify that first step.
So i noticed if we divide any p >= 5 by 6 we only get remainders of 1 or 5.
6 | 5 , r = 5
6 | 7 , r = 1
6 | 11, r = 5
6 | 13, r = 1
6 | 17, r = 5 and so on
so for my proof i am saying for p >= 5, p = 6k + 1 or 6k = 5
so for the first , p^2 + 2 = (6k+1)^2 + 2 = 36k^2 +12k + 3 = 3(12k^2 + 4k + 1)
and similarly, p^2 + 2 = (6k+5)^2 + 2 = 36k^2 + 60k + 27 = 3(12k^2 + 20k + 3)
so both of these are composite so that is great.
But is there a way i can justify that any prime p >= 5 can be written as 6k + 1 or 6k + 5?
I mean i can go through a ton of examples and the remainders are only 1 or 5 but examples don't mean anything.
Can someone help me justify that first step.
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