A problem of mathematical induction

Question

I am stuck by a problem of mathematical induction. The problem is:

Let $x$ and $y$ be integers. Prove that for each positive integer $n$ which is odd, we have $(x + y)$ divides $(x^n + y^n)$.

What I did is as follows:

For $n = 1$, $x^n + y^n = x + y$, which is obviously divisible by $x + y$;

For $n = k$ (odd), assume that $x^k + y^k$ is divisible by $x + y$, namely, \begin{equation} x^k + y^k = a(x+y), \tag{1} \end{equation} where $a$ is an integer;

For $n = k+2$, we have \begin{equation} x^{n} + y^{n} = x^{k+2} + y^{k+2}. \tag{2} \end{equation} I don't know how to proceed. How to make use of eq. (1) to prove the right-hand side of eq. (2) is divisible by $(x+y)$?

Answer 1

Note that:

\begin{align*} x^{k+2}+ y^{k+2} &= x^2(x^k + y^k - y^k) + y^{k+2} \\ &= x^2(x^k + y^k) + y^k(y^2-x^2)\\ &= x^2(x^k + y^k) + y^k(y-x)(x+y) \end{align*}

Answer 2

You can derive this directly from

• $x^n -y^n = (x-y)\sum_{k=0}^{n-1}x^{n-1-k}y^k$

by noting that $n= 2i+1$ and writing

$$x^{2i+1} + y^{2i+1} = x^{2i+1} - (-y)^{2i+1}$$

Answer 3

In context:

$x^k+y^k=a(x+y).$

$x^{k+2}= a(x+y)x^2-y^kx^2.$

$y^{k+2}=a(x+y)y^2-x^ky^2.$

$x^{k+2}+y^{k+2} = $

$a(x+y)(x^2+y^2)-y^2x^2(y^{k-2}+x^{k-2})$

First summand divisible by $(x+y)$.

Second summand: Divisible by (x+y) (hypothesis?).

Can I invoke the hypothesis for $k-2?$

Comments welcome.

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