Prove by induction that ${x}^{2k}-{y}^{2k}=(x+y)t$ for all natural numbers $k$ and $t$. Although it is an elementary algebraic result proved easily using algebra but I wanted to prove it inductively.
In my endeavor to do this, I've reached to the step which not only requires the preceding term but also the penultimate term to be true but in induction we only starts with the assumption of the previous term to be true. So, I'm left wondering how could I proceed further?
My proof goes like this:
$a(k)={x}^{2k}-{y}^{2k}$
$a(1)={x}^{2}-{y}^{2}=(x+y)(x-y)$
So, $a(1)$ is true.
Now, let us assume that $a(n)$ is true.
So, ${x}^{2n}-{y}^{2n}=(x+y)p$    for some polynomial $p$
Now, 
$a(n+1)={x}^{2n+2}-{y}^{2n+2}$
$={x}^{2n+2}-{y}^{2n+2}-{x}^{2}{y}^{2n}+{x}^{2}{y}^{2n}+{y}^{2}{x}^{2n}-{y}^{2}{x}^{2n}$
$={x}^{2}({x}^{2n}-{y}^{2n})+{y}^{2}({x}^{2n}-{y}^{2n})-{x}^{2}{y}^{2}({x}^{2n-2}-{y}^{2n-2})$
$=({x}^{2}+{y}^{2})({x}^{2n}-{y}^{2n})-{x}^{2}{y}^{2}({x}^{2n-2}-{y}^{2n-2})$
$=({x}^{2}+{y}^{2})((x+y)p)-{x}^{2}{y}^{2}({x}^{2n-2}-{y}^{2n-2})$
Now I'm stuck with ${x}^{2}{y}^{2}({x}^{2n-2}-{y}^{2n-2})$ divisibility by 11 that is the divisibility of $a(n-1)$ term by 11.
 A: Let $C(k)$ be the claim that there exists a polynomial $P_k(x,y)$ in $x$, $y$ such that $$x^{2k} - y^{2k} = (x+y)P_k(x,y).$$  Then a proof by induction might be as follows.  For $k = 0$, we have $$x^0 - y^0 = 0 = (x+y) P_0(x,y)$$ the trivial choice $P_0(x,y) = 0$.  For $k = 1$, we have $$x^2 - y^2 = (x+y)(x-y),$$ thus $P_1(x,y) = x-y$ establishes $C(1)$ as true.  So there exists a positive integer $m$ such that $C(k)$ is true for all $0 \le k \le m$.  Then consider $$x^{2(m+1)} - y^{2(m+1)} = (x^{2m} - y^{2m})(x^2 + y^2) - x^2 y^2 (x^{2(m-1)} - y^{2(m-1)}).$$  But since $k = m-1$ and $k = m$ both satisfy $0 \le k \le m$, we have $$x^{2m} - y^{2m} = (x+y)P_m(x,y) \\ x^{2(m-1)} - y^{2(m-1)} = (x+y)P_{m-1}(x,y),$$ hence $$x^{2(m+1)} - y^{2(m+1)} = (x+y)\left((x^2+y^2)P_m(x,y) - x^2 y^2 P_{m-1}(x,y)\right) = (x+y)P_{m+1}(x,y),$$ with the choice $$P_{m+1}(x,y) = (x^2+y^2)P_m(x,y) - x^2 y^2 P_{m-1}(x,y).$$  This establishes the induction step, hence claim $C(m+1)$ is also true.  Moreover, this proof gives us a recursion relationship for the polynomial $P_k(x,y)$, with the initial conditions $P_0(x,y) = 0$, $P_1(x,y) = x-y$.  We could use this to calculate $$P_2(x,y) = x^3 - x^2 y + x y^2 - y^3, \\ P_3(x,y) = x^5 - x^4 y + x^3 y^2 - x^2 y + x y^4 - y^5, \\ \text{etc.}$$  Of course, this suggests the explicit solution $$P_k(x,y) = \sum_{j=0}^{2k-1} x^{2k-1-j} (-y)^j,$$ and we could perform the multiplication $$\begin{align*} (x+y)P_k(x,y) &= (x+y)\sum_{j=0}^{2k-1} x^{2k-1-j} (-y)^j \\ &= \sum_{j=0}^{2k-1} x^{2k-j} (-y)^j - \sum_{j=0}^{2k-1} x^{2k-1-j} (-y)^{j+1} \\ &= \sum_{j=0}^{2k-1} x^{2k-j} (-y)^j - \sum_{j=1}^{2k} x^{2k-j} (-y)^j \\ &= x^{2k} (-y)^0 - x^{2k-2k} (-y)^{2k} \\ &= x^{2k} - ((-1)^2)^k y^{2k} \\ &= x^{2k} - y^{2k}. \end{align*}$$
