# Applications of Complex Number Analysis to Divisibility Problems : Two Undergrad Problems

## Problem

1. 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 $3$.
2. Prove that ${(x+y)}^n-x^n-y^n$ is divisible by $xy(x+y) \times {(x^2+xy+y^2)}^2$ if $n \equiv \pmod{6}$

## 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.