1

It is not hard to show that $a+b \ | \ a^n + b^n$ for odd $n$.

(because $f(x) = x^n - b^n = (x-b)h(x)$ we have $a - b \ | \ a^n - b^n$, so $a - (-b) \ | \ a^n - (-1)^n b^n$)

Is there a nice interpretation of this fact? (Or if not for all odd $n$, at least for $n = 3$ or $n = 5$?)

Krijn
  • 1,945
  • 16
  • 32
  • By interpretation do you want a proof? – Soham Oct 09 '15 at 16:09
  • What is the exact meaning of interpretation? – Soham Oct 09 '15 at 16:10
  • I gave a proof in the question, I would like a geometric or otherwise elementary interpretation of the result – Krijn Oct 09 '15 at 16:11
  • See also http://math.stackexchange.com/questions/188657/why-an-bn-is-divisible-by-a-b/1456649#1456649. – lhf Oct 09 '15 at 16:11
  • 1
    This is not an interpretation, but from the congruence point of view $b\equiv -a\pmod{a+b}$. So $b^n\equiv (-a)^n\equiv -a^n\pmod{a+b}$. – André Nicolas Oct 09 '15 at 16:15
  • I would like to note that it is easy to proof the result, but what I'm looking for is a visualization or interpretation of the result. I have a hard time to define 'interpretation' but Im looking for a explanation of this fact that makes it more clear. For example, if we see the right side as a sum of cubes, for n =3, then it is not intuitively clear why the left hand side divides it – Krijn Oct 09 '15 at 16:15
  • 1
    Sometimes a different proof is a way of giving an interpretation.... – Michael Burr Oct 09 '15 at 16:47
  • If $n$ is odd, then $a^n+b^n=(a+b)\left(a^{n-1}+a^{n-2}b+\cdots+b^{n-1}\right)$. – user236182 Oct 09 '15 at 16:58

2 Answers2

0

As you mentioned in the last comment you may see $a^3+b^3=(a+b)(a^2-ab+b^2)$

You may have a general view of this as $a^n+b^n$ has always $(a+b)$ as a factor.

You can prove this by binomial expansion.

Soham
  • 9,990
0

I have some interpretation of this fact, for $n = 3$ (and also other $n$ if you are willing to imagine $n$-cubes)

Because $f(x) = x^3 + a^3 = (x + a)(x + \zeta_3 a)(x + \zeta_3^2 a)$ we get that $a^3 + b^3 = (a+b)(\zeta_3 a+ b)(\zeta_3^2 a + b)$ where $\zeta_3 = -\frac{1}{2}-\frac{i \sqrt{3}}{2}$.

We can plot these in $\mathbb{C}$ in the following way: Plane with vectors

If we now rotate the vector $\zeta_3 a + b$ such that it aligns with the $i\mathbb{R}$ axis, and we rotate $\zeta_3^2 a + b$ such that it aligns with the axis perpendicular to the plane and then complete the cube, we get a cube with volume $a^3 + b^3$ with one rib being $a+b$.

It is possible to compute these rotated vectors, although too time consuming for me at this moment.

Krijn
  • 1,945
  • 16
  • 32