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Topic: Divisible by 42 (Read 2723 times) |
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PeterR
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Divisible by 42
« on: May 26th, 2013, 1:10pm » |
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Why is (x^7 - x) always divisible by 42?
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Grimbal
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Re: Divisible by 42
« Reply #1 on: May 26th, 2013, 2:17pm » |
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Because it is always divisible by 2, 3 and 7.
xp = x (mod p) for p prime.
x7 = x (mod 7)
x7 = x5 = x3 = x (mod 3)
x7 = x6 = ... = x2 = x (mod 2)
Therefore x7-x is a multiple of 7, 3 and 2, and therefore also of their product, 42.
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whizen
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Re: Divisible by 42
« Reply #2 on: May 29th, 2013, 2:36pm » |
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I didn't quite understand the divisibility by 3 and 2 proofs in Grimbal's reply. I have just been too lazy to find out, but not lazy enough to not work it out another way.
x7 - x = 0 (mod 7)
Hence, the original equation is divisible by 7
x7 - x = x ( x6 - 1)
x ( x3 - 1 )( x3 + 1 )
x ( x - 1 ) ( x2 + x + 1 ) ( x3 + 1 )
x ( x - 1 ) is divisible by 2
(x3 + 1) = (x + 1) mod 3...
hence ( x ( x - 1 ) (x3 + 1) ) {three consecutive numbers (mod 3)} is divisible by 3
Together, the product is divisible by 2*3*7 = 42
I would be delighted to know how Grimbal's divisibility tests by 2 & 3 work.
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towr
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Re: Divisible by 42
« Reply #3 on: May 29th, 2013, 10:16pm » |
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on May 29th, 2013, 2:36pm, whize wrote:| I would be delighted to know how Grimbal's divisibility tests by 2 & 3 work. |
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It follows the general rule he stated before, that for prime p: xp = x (mod p)
3 and 2 are primes.
So x7 = x3*x4, and then we can simplify the first factor x3*x4 = x*x4 (mod 3)
repeat a couple of times (for each = he wrote)
x*x4 = x3*x2 = x * x2 (mod 3)
x*x2 = x3 = x (mod 3)
Similar for 2.
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| « Last Edit: May 29th, 2013, 10:19pm by towr » |
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