1. ## Factorization induction proof.

Hi, I'm trying to proove the following by induction. I think I'm done, but how satisfactory would this proof be for a mathematician?

$\displaystyle x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$. If we set $\displaystyle n=1$ it is satisfied, so we suppose it is true for some $\displaystyle n \in \mathbb{N} > 1$. Then, for $\displaystyle n+1$

$\displaystyle x^{n+1}-y^{n+1}=xx^n-yy^n=x[(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})+y^n]-yy^n$
$\displaystyle =x(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})+xy^n-yy^n$
$\displaystyle =x(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})+y^n(x-y)$
$\displaystyle =(x-y)[(x^{n}+x^{n-1}y+...+x^{2}y^{n-2}+xy^{n-1})+y^n]$

2. Suppose

$\displaystyle x^n-y^n\ =\ (x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1})\ \ldots\ \fbox1$

is true. Multiplying through by $\displaystyle x$ gives

$\displaystyle x^{n+1}-xy^n\ =\ (x-y)(x^n+x^{n-1}y+\cdots+x^2y^{n-2}+xy^{n-1})\ \ldots\ \fbox2$

while multiplying $\displaystyle \fbox1$ by $\displaystyle y$ gives

$\displaystyle x^ny-y^{n+1}\ =\ (x-y)(x^{n-1}y+x^{n-1}y^2+\cdots+xy^{n-1}+y^n)\ \ldots\ \fbox3$

$\displaystyle \fbox2+\fbox3$ gives

$\displaystyle x^{n+1}+y^{n+1}+x^ny-xy^n\ =\ (x-y)(x^n+2\left[x^{n-1}y+x^{n-2}y^2+\cdots+x^2y^{n-2}+xy^{n-1}\right]+y^n)$

But

$\displaystyle x^ny-xy^n$

$\displaystyle =\ xy(x^{n-1}-y^{n-1})$

$\displaystyle =\ xy(x-y)(x^{n-2}+x^{n-3}y+\cdots+xy^{n-3}+y^{n-2})$

$\displaystyle =\ (x-y)(x^{n-1}y+x^{n-2}y^2+\cdots+x^2y^{n-2}+xy^{n-1})$

Hence the extra terms on both sides should cancel, giving the result we need.

Important note: Because we assumed the result to be true for both $\displaystyle n$ and $\displaystyle n-1$ in order to prove it true for $\displaystyle n+1,$ the whole proof by induction should be completed by showing that the result is true for both $\displaystyle n=1$ and $\displaystyle n=2.$ (Showing that it is true for just $\displaystyle n=1$ is not good enough.)

3. Hi, thanks for your post.

As you used that the result is true for $\displaystyle n-1$ it's necessary to show that it is true for both $\displaystyle n=2$ and $\displaystyle n=1$, but it will be in general for $\displaystyle n$,$\displaystyle n+1$ and $\displaystyle n+2$ for some $\displaystyle n \in \mathbb{N}$ isn't it?