• 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.

 

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

You May Also Like

Mathematical Logic – The basic introduction

What is Logic? If mathematics is regarded as a language, then logic is its grammar. In other words, logical precision has the same importance in mathematics as grammatical accuracy in a language. As linguistic grammar has sentences, statements— logic has them too. After we discuss about Sentence & Statements, we will proceed to further logical theories . Sentences & Statements…

Just another way to Multiply

Multiplication is probably the most important elementary operation in mathematics; even more important than usual addition. Every math-guy has its own style of multiplying numbers. But have you ever tried multiplicating by this way? Exercise: $ 88 \times 45$ =? Ans: as usual :- 3960 but I got this using a particular way: 88            45…

The ‘new’ largest known Prime Number

Great Internet Mersenne Prime Search (GIMPS) group has reported an all new Mersenne Prime Number (a prime number of type $2^P-1$) which is, now officially the largest prime number ever discovered. This number is valued to a whopping $2^{74207281}-1$ and contains 22,338,618 digits. It is quoted as M747207281 and is almost 5 million digits longer than the previous record holding prime number…

Albert Einstein and His introduction to the Concept of Relativity

Albert Einstein This name need not be explained. Albert Einstein is considered to be one of the best physicists in the human history. The twentieth century has undoubtedly been the most significant for the advance of science, in general, and Physics, in particular. And Einstein is the most luminated star of the 20th century. He literally created cm upheaval by…

Real Sequences

Sequence of real numbers A sequence of real numbers (or a real sequence) is defined as a function $ f: \mathbb{N} \to \mathbb{R}$ , where $ \mathbb{N}$ is the set of natural numbers and $ \mathbb{R}$ is the set of real numbers. Thus, $ f(n)=r_n, \ n \in \mathbb{N}, \ r_n \in \mathbb{R}$ is a function which produces a sequence…

Largest Prime Numbers

What is a Prime Number? An integer, say $ p $ , [ $ \ne {0} $ & $ \ne { \pm{1}} $ ] is said to be a prime integer iff its only factors (or divisors) are $ \pm{1} $ & $ \pm{p} $ . As? Few easy examples are: $ \pm{2}, \pm{3}, \pm{5}, \pm{7}, \pm{11}, \pm{13} $ …….etc.…