Prove using mathematical induction that $x^{2n} - y^{2n}$ is divisible by $x+y$ 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!

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 $\color{blue}{n\geq 1}$
