Prove using mathematical induction that $(x^{2n} - y^{2n})$ is divisible by $(x+y)$.
Step 1: Proving that the equation is true for $n=1 $
$(x^{2\cdot 1} - y^{2\cdot 1})$ is divisible by $(x+y)$
Step 2: Taking $n=k$
$(x^{2k} - y^{2k})$ is divisible by $(x+y)$
Step 3: proving that the above equation is also true for $(k+1)$
$(x^{2k+2} - y^{2k+2})$ is divisible by $(x+y)$.
Can anyone assist me what would be the next step? Thank You in advance!
Hint: Rewrite: $$x^{2k+2}-y^{2k+2}=(x^{2k+2}-y^{2k}x^2)+(y^{2k}x^2-y^{2k+2})=x^2(x^{2k}-y^{2k})+y^{2k}(x^2-y^2).$$
Added: Note it can be proved without induction: $$x^{2n}-y^{2n}=(x^2)^n-(y^2)^n=(x^2-y^2)(x^{2(n-1)}+x^{2(n-2)}y^2+\dots+x^2y^{2(n-2)}+y^{2(n-1)}),$$ and the first factor is divisible by $x+y$.
$$x^{2(k+1)}-y^{2(k+1)}=x^2x^{2k}-y^2y^{2k}=x^2x^{2k}\color{green}{-x^2y^{2k}+x^2y^{2k}}-y^2y^{2k}\\ =x^2(x^{2k}-y^{2k})+(x^2-y^2)y^{2k}.$$
The first term is divisible by $x+y$ by the induction hypothesis, and the second as well, by direct factorization.
Step 1: putting $n=1$, we get $$x^{2n}-y^{2n}=x^2-y^2=(x-y)(x+y)$$ above number is divisible by $(x+y)$. Hence the statement is true $n=1$
step 2: assuming that for $n=m$, $(x^{2n}-y^{2n})$ is divisible by $(x+y)$ then we have $$(x^{2m}-y^{2m})=k(x+y) \tag 1$$ Where, $k$ is an integer
step 3: putting $n=m+1$ we get $$x^{2(m+1)}-y^{2(m+1)}$$ $$=x^{2m+2}-y^{2m+2}$$ $$=x^{2m+2}-x^{2m}y^2-y^{2m+2}+x^{2m}y^2$$ $$=(x^{2m+2}-x^{2m}y^2)+(x^{2m}y^2-y^{2m+2})$$ $$=x^{2m}(x^2-y^2)+y^2(x^{2m}-y^{2m})$$ Setting the value of $(x^{2m}-y^{2m})$ from (1), we get $$x^{2m}(x-y)(x+y)+y^2k(x+y)$$ $$=k(x^{2m}(x-y)+y^2)(x+y)$$ It is clear that the above number is divisible by $(x+y)$. Hence the statement holds for $n=m+1$
Thus $(x^{2n}-y^{2n})$ is divisible by $(x+y)$ for all positive integers $\color{blue}{n\geq 1}$
For $n=k$, assume $P(k)$ is true, we have
$$x^{2k}-y^{2k}=A(x-y)$$, where A is a polynomial.
For $n=k+1$,
\begin{align} x^{2k+2}-y^{k+2}&=x^2[A(x-y)+y^{2k}]-y^{2k+2}\\&=A(x-y)x^2+x^2y^{2k}-y^{2k+2}\\&=A(x-y)x^2+y^{2k}(x^2-y^2)\\&=A(x-y)x^2+y^{2k}(x-y)(x+y)\\&=(x-y)[Ax^2+y^{2k}(x+y)]\\&=B(x-y)\text{, where } B \text{ is a polynomial}. \end{align}
When $n=1$,
$$x^{2n} - y^{2n} =x^2-y^2=(x+y)(x-y)$$
which is divisible by $(x+y)$. Assume true for $n=k$. Then $x^{2k}-y^{2k}$ is divisible by $x+y$. Putting $n=k+1$, $$x^{2n}-y^{2n}=x^{2k+2}-y^{2k+2}=\left(x^{k+1}-y^{k+1}\right) \left(x^{k+1}+y^{k+1}\right)$$
When $k$ is an odd number, $x+y$ is a factor of $x^{k+1}-y^{k+1}$. Similarly, when $k$ is an even number, $x+y$ is a factor of $x^{k+1} + y^{k+1}$. Hence $x^{2n}-y^{2n}$ is divisible by $x+y$ for integers $n>0$.
