Binomial theorem and odd power

Abstract

This paper deals with binomial and odd power $n=2m+1$. It presents two ways of grouping terms so that $(x+y{)}^n$ is always a sum of 2 coprime numbers: a first form $(x+y{)}^n=x{a}_n+y{b}_n$, and a second notable one with squares $(x+y{)}^n=x{c}_n^2+y{d}_n^2$ . Finally with $n$ prime, we show that $(a_n,b_n)$ prime factors are congruent to $1[2n]$, whereas $(c_n,d_n)$ congruent to $\pm 1[2n]$.

Introduction

We have searched how a powered number could systematically be shared into a sum of 2 coprime numbers. From binomial, we have studied different ways of grouping terms together so that $\mathrm{\forall }n,(x+y{)}^n=u_n+v_n,u_n\wedge v_n=1$ . With odd $n$ and $(x,y)$ coprime of opposite parity, we have found out two possibilities. They involve the same $f_n(x,y)$ functions that we must now introduce.

Definition

Let us define $f_n(x,y)$ functions as

Example

$\begin{array}{c}{f}_3(x,y)=x+3y\\ f_5(x,y)=x^2+10xy+5y^2\\ f_7(x,y)=x^3+21x^{2}y+35x{y}^2+7y^3\\ f_9(x,y)=x^4+36x^{3}y+126x^2{y}^2+84x{y}^3+9y^4\\ ...\end{array}$

Algebraic properties

Propositions

Proof

Binomial theorem gives:

$\mathrm{\forall }n,(x+y{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^{n-k}{y}^k$

Here $n$ is odd. So (1) is simply obtained by grouping together the odd power of $x$ and $y$ (2) is a consequence of (1).

Indeed it gives $(x+y{)}^n=x{f}_n(x^2,y^2)+y{f}_n(y^2,x^2)$

Thus by multiplying: $(x-y{)}^n(x+y{)}^n=x^2{f}_n(x^2,y^2{)}^2+y^2{f}_n(y^2,x^2{)}^2$

And finally $(x^2-y^2{)}^n=x^2{f}_n(x^2,y^2)+y^2{f}_n(y^2,x^2)$

Which leads to the proposition by replacing $(x^2,y^2)\to (x,y)$

Examples for (2):

$\begin{array}{c}(x-y{)}^3=x(x+3y{)}^2-y(y+3x{)}^2\\ (x+y{)}^3=x(x-3y{)}^2+y(y-3x{)}^2\end{array}$

$\begin{array}{c}(x-y{)}^5=x(x^2+10xy+5y^2{)}^2-y(y^2+10xy+5x^2{)}^2\\ (x+y{)}^5=x(x^2-10xy+5y^2{)}^2+y(y^2-10xy+5x^2{)}^2\end{array}$

Examples in  $\mathbb{Z}$:

$\begin{array}{ccc}17^3& =(5+12{)}^3=5\times 31^2& +12\times 3^2\\ 17^5& =(5+12{)}^5=5\times 145^2& +12\times 331^2\\ 17^7& =(5+12{)}^7=5\times 6929^2& +12\times 3767^2\\ 17^9& =(5+12{)}^9=5\times 138911^2& +12\times 42921^2\end{array}$

Proposition Template:Box

Proof

(1) implies (3)

(4):

$\begin{array}{cccc}(x+y{)}^n& -& (x-y{)}^n& =2y{f}_n(y^2,x^2)\\ u^n& -& v^n& =(u-v)(\sum _{k=0}^{2m}{u}^{2m-k}{v}^k)\end{array}$

So $f_n(y^2,x^2)=\sum _{k=0}^{2m}(x+y{)}^{2m-k}(x-y{)}^k$

Coprimality

Let us consider a more detailed form of $f_n(x,y)$ :

$f_n(x,y)=\underset{\text{k=0}}{\underbrace{x^m}}+\underset{\text{k=m}}{\underbrace{n{y}^m}}+\underset{\text{k=m-1}}{\underbrace{nmx{y}^{m-1}}}+\sum _{k=1}^{m-2}(\genfrac{}{}{0ex}{}{n}{2k})x^{m-k}{y}^k$

Proposition

Proof

First, $f_n(x,y)=x^m+n{y}^m+xyP(x,y)$ so $(x,y)$ of opposite parity implies $f_n(x,y)$ and $f_n(y,x)$ odd.

The rule on gcd, $a\wedge b=a\wedge b+ka$, immediately implies (6) and (7).

Indeed, $f_n(x,y)=x^m+yP(x,y)$.

And for $n\in \mathbb{P}$, $n\mid (\genfrac{}{}{0ex}{}{n}{2k})$ so $\frac{f_n(nx,y)}{n}=y^m+xP(x,y)$

Assertion (5) needs more attention.

Let us consider $p$ a common odd prime divisor.

The second form gives us $(x-y{)}^n\equiv 0[p]$, thus $x\equiv y[p]$

According to the definition of $f_n$

$f_n(x,y)\equiv \sum _{k=0}^m(\genfrac{}{}{0ex}{}{n}{2k})x^{m-k}{y}^k\equiv x^m(\sum _{k=0}^m(\genfrac{}{}{0ex}{}{n}{2k}))\equiv x^m(2^{m-1})[p]$

Thus $f_n(x,y)\equiv 0[p]\Rightarrow x\equiv 0[p]$, and the same $y\equiv 0[p]$

Every divisor of $f_n(x,y)$ and $f_n(y,x)$ divides $x$ and $y$

Prime factors

Propositions

Proofs here for (10) p=±1[2n] and (8) p=1[2n] on math.stackexchange.com/

Note

Fermat theorem gives $f_n(x,y)\equiv \pm 1[n]$ and $f_n(x^2,y)\equiv 1[n]$ . But what a surprise to discover that it also applies to all the prime factors! And much more specifically on the $f_n(x^2,y^2)$

Let us remind the Fermat's theorem on sums of two squares: $x^2+y^2=\prod p_i^{v_i}\Rightarrow p_i\equiv 1[4]$

And the Euler's theorem: $x^2+3y^2=\prod p_i^{v_i}\Rightarrow p_i\equiv 1[3]$ , which is here $f_3(x^2,y^2)$

Fermat had discovered that $2^n-1$ and $\frac{2^n+1}{3}$ had $1[2n]$ prime factors (cf letters to Mersenne and Frenicle in 1640)

Let us note that these $f_n(x^2,y^2)$ also appear in Fermat-Wiles theorem with (3)

Examples for $f_n(x,y)\equiv \pm 1[2n]$

$\begin{array}{cccc}{f}_5(6,11)& =1301& \equiv 1& [10]\\ f_7(6,11)& =181\times 239& \equiv -1\times 1& [14]\\ f_{11}(6,11)& =47820079& \equiv 1& [22]\\ f_{13}(6,11)& =8969\times 177269& \equiv -1\times 1& [26]\\ f_{19}(6,11)& =151\times 386989747169& \equiv -1\times -1& [38]\\ f_{23}(6,11)& =5278223\times 12238317893& \equiv -1\times -1& [46]\\ f_{29}(6,11)& =233\times 521\times 1309699\times 14932891583& \equiv 1\times -1\times 1\times -1& [58]\\ f_{31}(6,11)& =683\times 8803\times 13128774105430771& \equiv 1\times -1\times 1& [62]\\ ...\end{array}$

Examples for $f_n(x^2,y)\equiv 1[2n]$ . The number of $-1[n]$ factors is even

$\begin{array}{cccc}{f}_5(6^2,11)& =5861& \equiv 1& [10]\\ f_7(6^2,11)& =507809& \equiv 1& [14]\\ f_{11}(6^2,11)& =89\times 6029\times 7129& \equiv 1\times 1\times 1& [22]\\ f_{13}(6^2,11)& =883\times 2341\times 160627& \equiv -1\times 1\times -1& [26]\\ f_{17}(6^2,11)& =67\times 187067\times 199591969& \equiv -1\times -1\times 1& [34]\\ f_{19}(6^2,11)& =229\times 683\times 1901\times 2963\times 246469& \equiv 1\times -1\times 1\times -1\times 1& [38]\\ f_{23}(6^2,11)& =367\times 4457595074737380607& \equiv -1\times -1& [46]\\ f_{29}(6^2,11)& =1069838719517673460520580221& \equiv 1& [58]\\ f_{31}(6^2,11)& =92861463243343352659695472649& \equiv 1& [62]\\ ...\end{array}$

Examples with both squared variables:$f_n(x^2,y^2)\equiv 1[2n]$

$\begin{array}{cccc}{f}_5(6^2,11^2)& =118061& \equiv 1& [10]\\ f_7(6^2,11^2)& =34188379& \equiv 1& [14]\\ f_{11}(6^2,11^2)& =23\times 89\times 2377\times 586961& \equiv 1\times 1\times 1\times 1& [22]\\ f_{13}(6^2,11^2)& =30187\times 27342279823& \equiv 1\times 1& [26]\\ f_{19}(6^2,11^2)& =2129\times 3079\times 3039225397425209& \equiv 1\times 1\times 1& [38]\\ f_{23}(6^2,11^2)& =1663964075070633572800332299& \equiv 1& [46]\\ f_{29}(6^2,11^2)& =59\times 11250493\times 60508154689065340703592763& \equiv 1\times 1\times 1& [58]\\ f_{31}(6^2,11^2)& =27466437659\times 422603394089373421296083801& \equiv 1\times 1& [62]\\ ...\end{array}$