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Suppose I have a simple polynomial in $\{a,b\}$, defined as $a^k-b^k \ \forall \ k\in \mathbb{Z}^{\geq0}$. If I know one of the factor is $(a-b)$, is there a way to get a representation of its remaining factors in Wolfram Language?

I know one of its representation is $\sum_{j=0}^{k-1}{(a^{(k-1)-j}\ b^j)}$ and Mathematica recognizes that

Sum[a^((k-1)-j) b^j,{j,0,k-1}]

gives

(a^k - b^k)/(a - b)

But if I ask

FullSimplify[(a^k-b^k)/(a-b),Assumptions->k\[Element]NonNegativeIntegers]

It is unable to do anything.

Also what is the proper way to give assumptions?

Is the above expression intepreted differently if I give as

Assuming[k\[Element]NonNegativeIntegers,FullSimplify[(a^k-b^k)/(a-b)]]

Also tried factoring directly without giving a single factor to no success,

Assuming[k\[Element]NonNegativeIntegers,Factor[a^k-b^k]]
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1 Answer 1

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It is well known that $x^k - 1$ factors into the cyclotomic polynomials of the divisors of $k$. To factor $a^k - b^k$ instead, we take the cyclotomic polynomials in $a/b$ and remove the denominator.

For Mathematica to give you a general expression in k, it would have to generate an infinite product of (...)^Boole[Divides[k, d]] terms, or quantify over the Divisors, which seems beyond what Factor is capable of (all the documentation examples result in polynomial expressions).

factorDiffOfPows[k_Integer] :=  (Expand[Numerator@Together@Cyclotomic[#, a/b]])& /@ Divisors@k

Print[factorDiffOfPows@10]
Print[AllTrue[Range@105, Expand[Times @@ factorDiffOfPows@#] == a^# - b^#&]]
(*
{a - b, a + b, a^4 + a^3*b + a^2*b^2 + a*b^3 + b^4, a^4 - a^3*b + a^2*b^2 - a*b^3 + b^4}
True
*)

Try it online!

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