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

$(a^n - b^n)$

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?

## closed as too broad by user223391, 6005, user147263, rogerl, Carl MummertAug 31 '15 at 2:34

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• So you're asking us to prove six different things in one question? – Git Gud Jan 17 '14 at 9:33
• Consider it a polynomial in $a$. Use the factor/remainder theorem. – Macavity Jan 17 '14 at 9:56
• I voted to put this on hold because it is unclear. Are $a$ and $b$ numbers, or polynomial variables? – Carl Mummert Aug 31 '15 at 2:34
• @Ramit $\,$ Hint $\,\ {\rm mod}\ a+b\!:\,\ b\equiv -a\,\Rightarrow\, a^n+b^n\equiv a^n+(-a)^n\ \$ – Bill Dubuque Jun 16 '17 at 19:42
• try for a=3,b=1,n=3 you will see (a-b) divides a^n+b^n – Anukul gaurav Aug 3 '18 at 20:13

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 :

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

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

• actually I was looking for an intuition based answer. your steps are appearing too technical for me. – Ramit Jan 17 '14 at 9:39
• @Ramit, how about the second line? – lab bhattacharjee Jan 17 '14 at 9:41
• @Ramit: If you find symbols unfamiliar, you can just substitute in small values for $n$ starting 1,2,3,... to see what intuitively the proof is doing. For example, $a^2-b^2=(a-b)...$. – user21820 Jan 17 '14 at 9:42

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: $$a^n+b^n = (a+b)\left(\sum_{k=0}^{n-1} (-1)^k a^{n-1-k}b^k\right) +(1+(-1)^n)b^n$$

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

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.