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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 \$

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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## Hard drive – Time Machine would like to secure 4.50 PB 266.14 GB

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.

• 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?