# Applications of Complex Number Analysis to Divisibility Problems

- Prove that $ {(x+y)}^n-x^n-y^n$ is divisible by $ xy(x+y) \times (x^2+xy+y^2)$ if $ n$ is an odd number not divisible by $ 3$ .
- Prove that $ {(x+y)}^n-x^n-y^n$ is divisible by $ xy(x+y) \times {(x^2+xy+y^2)}^2$ if $ n \equiv \pmod{6}1$

**Solution**

1.Considering the given expression as a polynomial in $ y$ , let us put $ y=0 $ . We see that at $ y=0 $ the polynomial vanishes (for any $ x$ ). Therefore our polynomial is divisible by $ y$ . Similarly, it is divisible by $ x$ as well. Thus the polynomial is divisible by $ xy$ .

To prove that it is divisible by $ x+y $ , put $ x+y=0 \ {or} \ y=-x $ . It is evident that for odd n we have : $ {(x+(-x)}^n-x^n-{(-x)}^n = 0 $ for $ y=-x $ .

Consequently, our polynomial is divisible by $ x+y $ . It only remains to prove the divisibility of the polynomial by $ x^2 +xy+y^2$ , which also be written as $ (y-x\epsilon)(y-x{\epsilon}^2 ) $ where $ \epsilon^2+\epsilon+1=0 $ .

For this purpose it only remains to replace $ y $ first by $ x \epsilon $ and then by $ x\epsilon^2 $ to make sure that with these substitutions the polynomial vanishes. Since, by hypothesis, $ n$ is not divisible by 3, it follows that $ n=3l+1 \ or \ 3l+2 $ , for every $ l \in \mathbb{Z} $ , in which $ 3l+1$ is not acceptable since $ n$ is odd from the problem. At $ y=x\epsilon $ the polynomial attains the following value

$ {(x+x\epsilon)}^n-x^n-{(x\epsilon)}^n=x^n [{(1+\epsilon)}^n-1-\epsilon^n] \\ =x^n {(-\epsilon^2)}^n -1 -\epsilon^n …. $ since ($ 1+\epsilon + \epsilon^2=0 $ ) substituting $ n=3l+2 $ we get

$ 1+\epsilon+\epsilon^2 =0 $

Likewise we prove that at $ y=x\epsilon^2$ the polynomial vanishes as well, and consequently, its by divisibility by $ xy(x+y) \times (x^2+xy+y^2) $ is proved.

2.To prove the second statement, let us proceed as follows. Let the quantities $ {-x, -y, \, and \, x+y} $ be the roots of a cubic equation $ X^3-rX^2-pX-q=0 $ . Then by virtue of the known relations between the roots of an equation and its coefficients we have $ r=-x-y-(x+y)=0 \\ -p=xy-x(x+y)-y(x+y)$ or $ p=x^2+xy+y^2$ and $ q=xy(x+y)$ .

Thus, $ -x, \, -y \, x+y$ are the roots of the equation $ X^3-pX-q=0 $ where $ p=x^2+xy+y^2$ and $ q=xy(x+y) $

Put $ {(-x)}^n-{(-y)}^n+{(x+y)}^n=S_n$ . Among successive values of $ S_n$ , there exist the relationship $ S_{n+3}=pS_{n+1}+qS_n$ ,: $ S_1$ being equal to zero.

Let us prove that $ S_n$ is divisible by $ p^2$ if $ n \equiv 1 \pmod{6}$ using the method of mathematical induction. Suppose $ S_n $ is divisible by $ p^2 $ and prove that then $ S_{n+6} $ is also divisible by $ p^2$ .

So, using this relation we get that

$ S_{n+6}=p(pS_{n+2} + qS_{n+1}) + q(pS_{n+1}+qS_n) \\ =p^2S_{n+2}+2pqS_{n+1}+q^2S_n$ .

Since, by supposition, $ S_n$ is divisible by $ p^2$ , it suffices to prove that $ S_{n+1}$ is divisible by $ p$ . Thus we only have to prove than $ S_n={(x+y)}^n+(-x)^n+(-y)^n$ is divisible by $ p=x^2+xy+y^2$ if $ n \equiv 2 \pmod{6}$ , we easily prove our assertion. And so, assuming that $ S_n$ is divisible by $ p^2$ , we have proved that (from induction) $ S_{n+6}$ is also divisible by $ p^2$ . Consequently $ S_n ={(x+y)}^n+(-x)^n+(-y)^n={(x+y)}^n-x^n-y^n$ for any $ n \equiv 1 \pmod{6}$ is divisible by $ p^2={(x^2+xy+y^2)}^2 $ .

Now it only remains to prove its divisibility by $ x+y $ and by $ xy$ , which is quite elementary.

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