Proof of $a^n+b^n$ divisible by $a+b$ when $n$ is odd [closed]

Solution 1:

If $a=b, a^n=b^n$

or $\displaystyle a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$

If $a=-b, a^{2m+1}=(-b)^{2m+1}=-b^{2m+1}$

Induction :

$\displaystyle a^n-b^n=a(a^{n-1}-b^{n-1})+b^{n-1}(a-b) $

$\displaystyle a^{2m+1}+b^{2m+1}=a^2(a^{2m-1}+b^{2m-1})-b^{2m-1}(a^2-b^2) $

Solution 2:

As you already grasp $\;a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1})$. Now, if $\;n\;$ is odd then $\;b^n=-(-b)^n$, so using the above

$$a^n+b^n=a^n-(-b)^n=(a-(-b))(a^{n-1}+a^{n-2}(-b)+\ldots+a(-b)^{n-2}+(-b)^{n-1})=$$

$$=(a+b)(a^{n-1}-a^{n-2}b+\ldots-ab^{n-2}+b^{n-1})$$

Since $\;(-b)^n=b^n\iff n\;$ is even. For example, $\;(-b)^{n-2}=-b^{n-2}.$

Solution 3:

for $a^n+b^n$, to treat the odd and even cases in a single framework-- if you do the long division you will notice the following identity: $$ a^n+b^n = (a+b)\left(\sum_{k=0}^{n-1} (-1)^k a^{n-1-k}b^k\right) +(1+(-1)^n)b^n $$