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.

 

Default image
Gaurav Tiwari
Gaurav Tiwari is a professional graphic & web designer from New Delhi, India. gauravtiwari.org is his personal space where he writes on blogging, digital marketing, content writing, learning and business growth. Gaurav has contributed in developing more than 325 brands worldwide and while you are reading this, he's busy building a couple more.

Newsletter Updates

Enter your email address below to subscribe to our newsletter

Leave a Reply