Tag: PB

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?