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.....all prime numbers larger than 3 are 6n+1 or 6n-1? E.g 73 = 6x12+1.
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Yeah, it's not even difficult to prove this, which doesn't make it not a cool fact! First, note that prime numbers (bigger than 3) can't be divisible by six, since then they wouldn't be prime. Nor could they have a remainder of 2 or 4 when divided by six, since then they would be even. So the only other remainders possible are 1,3,5. But if it's divisible by six,...
13:57 Mon 02nd Jan 2023
Yeah, it's not even difficult to prove this, which doesn't make it not a cool fact!
First, note that prime numbers (bigger than 3) can't be divisible by six, since then they wouldn't be prime. Nor could they have a remainder of 2 or 4 when divided by six, since then they would be even. So the only other remainders possible are 1,3,5. But if it's divisible by six, then it's also divisible by three, and adding three to it (ie, the remainder 3 option) would *still* make it divisible by three. Hence the only remainder possible is 1 or 5, or as you say 6n-1 and 6n+1.
Also, it follows from this that all prime numbers squared are one more or less than a multiple of 24:
(6n-1)^2 = 36n^2 - 12n + 1 = 12 (3n^2-n)+1
(6n+1)^2 = 36n^2 + 12n + 1 = 12 (3n^2+n)+1
showing that the squares are one more or less than a multiple of 12. But 3n^2 is even if n is even, and odd if n is odd, so that 3n^2 + n and 3n^2 - n are both always even.
Hence, all squares of primes are one more or less than a multiple of 12*2 = 24.
Fun observation TTT :)
First, note that prime numbers (bigger than 3) can't be divisible by six, since then they wouldn't be prime. Nor could they have a remainder of 2 or 4 when divided by six, since then they would be even. So the only other remainders possible are 1,3,5. But if it's divisible by six, then it's also divisible by three, and adding three to it (ie, the remainder 3 option) would *still* make it divisible by three. Hence the only remainder possible is 1 or 5, or as you say 6n-1 and 6n+1.
Also, it follows from this that all prime numbers squared are one more or less than a multiple of 24:
(6n-1)^2 = 36n^2 - 12n + 1 = 12 (3n^2-n)+1
(6n+1)^2 = 36n^2 + 12n + 1 = 12 (3n^2+n)+1
showing that the squares are one more or less than a multiple of 12. But 3n^2 is even if n is even, and odd if n is odd, so that 3n^2 + n and 3n^2 - n are both always even.
Hence, all squares of primes are one more or less than a multiple of 12*2 = 24.
Fun observation TTT :)
The Heegner Numbers are my favourites. Go exploring from here and some remarkable facts will emerge. Almost An Integer is fascinating.
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oh god I was gonna say 2 divides 6n+2 , 4 , and 3 divides 6n+3.
if anyone is serious - Elementary Number Theory David Burton ( OU set text for a defunct course ( I did it in 1986) - some kind soul has put all the worked solutions on the internet.
So you can kinda do it yourself, and when you get stumped look up the answer. Elementary here means anything less complex than eulers totient func
and yes I wrote and thanked him
if anyone is serious - Elementary Number Theory David Burton ( OU set text for a defunct course ( I did it in 1986) - some kind soul has put all the worked solutions on the internet.
So you can kinda do it yourself, and when you get stumped look up the answer. Elementary here means anything less complex than eulers totient func
and yes I wrote and thanked him
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