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

Question

I read somewhere that

$(a^n - b^n)$

1. It is always divisible by $a-b$.
2. When $n$ is even it is also divisible by $a+b$.
3. When $n$ is odd it is not divisible by $a+b$.

and

$(a^n + b^n)$

1. It is never divisible by $a-b$.
2. When $n$ is odd it is divisible by $a+b$.
3. When $n$ is even it is not divisible by $a+b$.

I wonder what's the proof for this.

First postulate is clear. $(a-b)$ would or would not be a factor. Any light on others?

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

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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) $

Answer 2

If the first postulate is clear, then $a^\text{even}-b^\text{even}=a^{2n}-b^{2n}=(a^2)^n-(b^2)^n$, which is divisible
through $(a^2-b^2)$. But $a^2-b^2=(a-b)(a+b)$. It should then be obvious why this trick can
not be applied for odd values of the exponent, implying $(3)$. As far as the second group is con-
cerned, try giving small values to n, such as $3$ or $5$, and compute $(a+b)(a^2-ab+b^2)$ or
$(a+b)(a^4-a^3b$$+a^2b^2-ab^3+b^4)$. Notice how the terms just cancel each other out, due to
sign-alternation, and all that remains is only the first and the last. Then try applying a similar
trick to even values of n, and notice how that would force the greatest powers of a and b to be
of opposite signs.

Answer 3

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, $\;(-b)^{n-2}=-b^{n-2}.$