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# Tag: PB

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## Number Theory – Prove the highest power of $ (a ^ 2-ab + b ^ 2) $, which divides $ a ^ p-b ^ p- (a-b) ^ p $ for odd primes $ n $

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It's obvious that odd $ n in Bbb N $, $ a ^ n-b ^ n- (a-b) ^ n $ is divisible by $ ab (a-b) $ (With $ n = 1 $ a special case in which $ a ^ n-b ^ n- (a-b) ^ n $ is zero). This can be considered as fact with respect to integers, but is stronger with respect to these polynomials.

It also seems that for prime $ p> 3 $, it is also divisible by $ p $ and there is one more simple factor

$$

Left. (a ^ 2-ab + b ^ 2) right | a ^ p -b ^ p- (a-b) ^ p

$$

and especially if $ p = 6k-1 $ for some $ k in Bbb N $ then $ a ^ p -b ^ p- (a-b) ^ p $ is divisible by $ (a ^ 2-ab + b ^ 2) $ but not through $ (a ^ 2-ab + b ^ 2) ^ 2 $; while if $ p = 6k + 1 $ then $ a ^ p -b ^ p- (a-b) ^ p $ is divisible by $ (a ^ 2-ab + b ^ 2) ^ 2 $,

This property does not depend on $ p $ Be Prime: The remainder of the statement is when it's odd $ n $ is divisible by $ 3 $ then $ a ^ n-b ^ n- (a-b) ^ n $ is not divisible by $ (a ^ 2-ab + b ^ 2) $,

I want to prove that to everyone $ k in Bbb N $

$$

a ^ {6k + 1} -b ^ {6k + 1} – (ab) ^ {6k + 1} = (6k + 1) ab (ab) (a ^ 2-ab + b ^ 2) ^ 2 P_1 ( a, b)

$$

for a polynomial $ P_1 (a, b) in Bbb Z[a,b]$, and

$$

a ^ {6k + 5} -b ^ {6k + 5} – (ab) ^ {6k + 5} = (6k + 5) ab (ab) (a ^ 2-ab + b ^ 2) P_5 (a, b)

$$

for a polynomial $ P_5 (a, b) in Bbb Z[a,b]$ the self is not a multiple of $ (a ^ 2-ab + b ^ 2) $, and

$$

a ^ {6k + 3} -b ^ {6k + 3} – (a-b) ^ {6k + 3} = (6k + 3) a b (a-b) P_3 (a, b)

$$ from where $ P_3 (a, b) $ is not divisible by $ (a ^ 2-ab + b ^ 2) $,

In trying to prove it, I noticed that $ (a ^ 2-ab + b ^ 2) = (a- omega b) (a – omega ^ 2) b $ from where $ omega $ is a non-trivial cube root of the unit. I had hoped that this would shed a light on why a multiple of $ 3 $ would not have the divisibility, but I can not see how that follows.

Is there a more powerful technique with which the given statements "fall out" or at least are easier to prove?

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Currently, you can not back up through Time Machine because backups fail with the following error message:

"Time Machine could not complete backing up to Time Machine." The backup hard drive requires 4.50 PB for backup, but only 337.69 GB is available. Select a larger backup disk or shrink the backup by excluding files. "

When I go to Time Machine Options, the estimated size of a full backup is 266.14 GB. The source volume is only 500 GB.

Things I've already tried:

- Run the First Aid Disk Utility on both the startup disk (the drive to be backed up) and the Time Machine diskette
- Restart the Mac
- The Time Machine disk is included in "Excluding Elements of Backups." It is.

Does anyone know why Time Machine could be? *solid* Overestimation of the size required for the backup?

4.50 PB is 4.500 Terabytes, right?

Hi Guys,

Does anyone know of any good places to buy Pbn posts? I have used https://www.wraithpbns.com/ and http://pbnlinkbuilding.com/. are both ok, more places that people have in their arsenal?

cheers

Graham

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