Write $x^k - y^k$ as the product of two factors.

Question

Write $x^k - y^k$ as the product of two factors.

$x^3 - y^3 = (x-y)(x^2+xy+y^2)$

$x^4 - y^4 = (x-y)(x^3+x^2y+xy^2+y^3)$

$x^5 - y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$

$x^6 - y^6 = (x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5)$

$x^k - y^k = (x-y)(x^{k-1}+\underbrace{x^{k-2}y + \cdots + xy^{k-2}}_{k-2} +y^{k-1})$

It's not exact what the middle terms should be, so I'm not sure how to write the expression for the second factor. I think I may need to use the parity of $k$ and some mention of mirroring, though mirroring sounds wrong.

Answer 1

Here is a variation without using induction. We want to use the sigma symbol and start with an example for small $n=4$ to better see what's going on.

We have \begin{align*} x^4-y^4&=(x-y)(x^3+x^2y+xy^2+y^3)\\ &=(x-y)(x^3y^0+x^2y^1+x^1y^2+x^0y^3)\tag{1}\\ &=(x-y)\sum_{k=0}^3 x^{3-k}y^k \end{align*}

Since we can write $x^3=x^3y^0$ and $y^3=x^0y^3$ each term in (1) has a representation $x^{3-k}y^k$, $0\leq k\leq 3$ which can be conveniently written using the sigma symbol $\sum$.

For the general case with integral $n>0$ we have \begin{align*} \color{blue}{x^n-y^n}&=(x-y)(x^{n-1}+x^{n-2}y+\cdots + xy^{n-2}+y^{n-1})\\ &=(x-y)(x^{n-1}y^0+x^{n-2}y^1+\cdots + x^1y^{n-2}+x^0y^{n-1})\\ &\,\,\color{blue}{=(x-y)\sum_{k=0}^{n-1} x^{n-1-k}y^k}\tag{2} \end{align*}

We prove (2) algebraically. We obtain for integral $n>0$ by starting with the right-hand side of (2) \begin{align*} \color{blue}{(x-y)\sum_{k=0}^{n-1} x^{n-1-k}y^k}&=x\sum_{k=0}^{n-1} x^{n-1-k}y^k-y\sum_{k=0}^{n-1} x^{n-1-k}y^k\tag{3.1}\\ &=\sum_{k=0}^{n-1} x^{n-k}y^k-\sum_{k=0}^{n-1} x^{n-1-k}y^{k+1}\tag{3.2}\\ &=\sum_{k=0}^{n-1} x^{n-k}y^k-\sum_{k=1}^{n} x^{n-k}y^{k}\tag{3.3}\\ &\,\,\color{blue}{=x^n-y^n} \end{align*} and the claim follows.

Comment:

• In (3.1) we multiply out $(x-y)\sum \Box=x\sum \Box - y\sum\Box$.

• In (3.2) we simplify by multiplying with $x$ resp. $y$. Note, we have $x\sum x^{n-1-k}y^k=\sum x^{1+(n-1-k)}y^k=\sum x^{n-k}y^k$.

• In (3.3) we shift the index $k$ in the right-hand side by one. We so have lower bound $k=1$ and upper bound $k=n$. We compensate this shift by substituting $k\to k-1$ within the sum. We now have the convenient situation that we have a telescoping sum with alike terms in the left-hand sum and the right-hand sum for $k=1,\ldots,n-1$, so that everything cancels out besides $x^n-y^n$.

Answer 2

The sum you bracketed can be expressed as $$x^{k-2}y+\cdots +xy^{k-2} =\sum_{i=1}^{k-2}x^{k-2+1-i}y^i =\sum_{i=0}^{k-1}x^iy^{k-j-1} -x^{k-1}-y^{k-1}$$ For a formal proof: Notice that $$a^2-b^2 = a^2+ab-ab-b^2 = a(a+b)-b(a+b) = (a-b)(a+b)$$ so that our result holds for $n=2$.Now suppose that $$a^k -b^k = (a-b)(a^{k-1}+ba^{k-2}+b^2a^{k-3}+b^3a^{k-4} +\cdots +b^{k-3}a^2 +b^{k-2}a +b^{k-1})$$ for some $k\in \mathbf{N}$ with $k\geqslant 2.$ Then \begin{align*} a^{k+1}-b^{k+1} & = aa^k-bb^k \\ &= aa^k-ba^k+ba^k-bb^k \\ &= a^k(a-b)+b(a^k-b^k) \\ &= a^k(a-b)+b\left((a-b)(a^{k-1}+ba^{k-2}+\cdots +b^{k-2}a +b^{k-1})\right) \hspace{8mm}\text{Inductive Hypothesis}\\ &= (a-b)(a^k +b(a^{k-1}+ba^{k-2}+\cdots +b^{k-2}a +b^{k-1}))\\ &= (a-b)(a^{k}+ba^{k-1}+b^2a^{k-2}+b^3a^{k-3} +\cdots +b^{k-2}a^2 +b^{k-1}a +b^{k}). \end{align*} Therefore, by the Principles of Mathematical Induction, we have that $$a^n -b^n = (a-b)(a^{n-1}+ba^{n-2}+b^2a^{n-3}+b^3a^{n-4} +\cdots +b^{n-3}a^2 +b^{n-2}a +b^{n-1})$$ for any $a$, $b\in \mathbf{R}$ and $n\in \mathbf{N}.\blacksquare$