($a^n$+$b^n$) is divisible by ($a$+$b$) when $n$ is odd (proof by induction)

Question

I am trying to prove that ($a^n$+$b^n$) is divisible by ($a$+$b$) when $n$ is odd.

I start off by checking for $n$ = $1$, and find that it is true for $n$ = $1$

Now I assume that it is true for $n$ = ($2k$-$1$),

i.e [$a^{2k-1}$+$b^{2k-1}$] is divisible by ($a$+$b$)

Finally, I proceed to checking for $n$ = ($2k$+$1$) :

$=$ {$a^{2k+1}$+$b^{2k+1}$}

$=$ {$a^2$$a^{2k-1}$+$b^2$$b^{2k-1}$}

Now I don’t understand how to prove that {$a^2$$a^{2k-1}$+$b^2$$b^{2k-1}$} is divisible by ($a$+$b$) based on my assumption earlier.

Appreciate any help

Answer 1

For the induction step in this proof to work, we need to check two base cases: $n = 1$ is obvious as you said, and for $n=3$:

$$ a+b\ |\ (a+b)(a^2-ab+b^2) = a^3+b^3$$

Since we proved two base cases, now we can assume it is true for $n=2k-3$ and $n=2k-1$, that means:

$$a+b\ |\ a^{2k-3}+b^{2k-3}\ \wedge\ a+b\ |\ a^{2k-1}+b^{2k-1}$$

Therefore:

$$a+b\ |\ (a^2+b^2)(a^{2k-1}+b^{2k-1})-a^2b^2(a^{2k-3}+b^{2k-3}) = a^{2k+1}+b^{2k+1}$$

we get that

$$a+b \ | \ a^{2k+1}+b^{2k+1}$$

Answer 2

Hint:

$$a^{2k+1}+b^{2k+1}=a^2(a^{2k-1}+b^{2k-1})-b^{2k-1}(a^2-b^2)$$