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 $(x+y)^{n}=x\cdot a_{n}+y\cdot b_{n}$, and a second notable one with squares $(x+y)^{n}=x\cdot c_{n}^{2}+y\cdot d_{n}^{2}$ . This latter form with $(c_{n},d_{n})=1$ is all the more remarkable given that it is unique when $x+y$ is prime . 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 $\forall n,~(x+y)^{n}=u_{n}+v_{n},~u_{n}\wedge v_{n}=1$ . 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

$\quad n=2m+1,~f_{n}(x,y)=\sum _{k=0}^{m}{n \choose 2k}x^{m-k}y^{k}$

Example

$\quad {\begin{array}{lll}f_{3}(x,y)=x+3y\\f_{5}(x,y)=x^{2}+10xy+5y^{2}\\f_{7}(x,y)=x^{3}+21x^{2}y+35xy^{2}+7y^{3}\\f_{9}(x,y)=x^{4}+36x^{3}y+126x^{2}y^{2}+84xy^{3}+9y^{4}\\...\end{array}}$

Algebraic properties

Propositions

$\quad {\begin{array}{lll}(x-y)^{n}&=xf_{n}(x^{2},y^{2})&-&yf_{n}(y^{2},x^{2})&\quad (1)\\(x-y)^{n}&=xf_{n}(x,y)^{2}&-&yf_{n}(y,x)^{2}&\quad (2)\end{array}}$

Proof

Binomial theorem gives:

$\forall n,~(x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}$

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 $(x+y)^{n}=xf_{n}(x^{2},y^{2})+yf_{n}(y^{2},x^{2})$

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 $(x^{2},y^{2})\rightarrow (x,y)$

Examples for (2):

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

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

Examples in  $\mathbb {Z}$:

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

Proposition

$\quad {\begin{array}{rll}(x+y)^{n}+(x-y)^{n}&=2xf_{n}(x^{2},y^{2})&\quad (3)\\x^{n}+y^{y}&={\frac {1}{2^{n-1}}}(x+y)f_{n}\left((x+y)^{2},(x-y)^{2}\right)&\quad (3bis)\end{array}}$

Proof

(1) implies (3)

(3bis) using $f_{n}(a/2,b/2)=f_{n}(a,b)/2^{m}$

Coprimality

Let us consider a more detailed form of $f_{n}(x,y)$ :

$\quad f_{n}(x,y)=\underbrace {x^{m}} _{\text{k=0}}+\underbrace {ny^{m}} _{\text{k=m}}+\underbrace {nmxy^{m-1}} _{\text{k=m-1}}+\sum _{k=1}^{m-2}{n \choose 2k}x^{m-k}y^{k}$

Proposition

$2\mid xy,~x\wedge y=1\Rightarrow {\begin{cases}f_{n}(x,y)\wedge f_{n}(y,x)=1&(5)\\n\nmid x\Rightarrow x\wedge f_{n}(x,y)=1&(6)\\n\mid x\Rightarrow x\wedge f_{n}(x,y)=n&(7)\end{cases}}$

Proof

First, $f_{n}(x,y)=x^{m}+ny^{m}+xyP(x,y)$ 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 $n\in \mathbb {P}$, $n\mid {n \choose 2k}$ so ${\dfrac {f_{n}(nx,y)}{n}}=y^{m}+xP(x,y)$

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 $f_{n}$

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

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 $y$

Examples: previous examples with prime factors

${\begin{array}{lll}17^{3}&=(5+12)^{3}=5\times (31)^{2}&+12\times (3)^{2}\\17^{5}&=(5+12)^{5}=5\times (5\cdot 29)^{2}&+12\times (331)^{2}\\17^{7}&=(5+12)^{7}=5\times (13^{2}\cdot 41)^{2}&+12\times (3767)^{2}\\17^{9}&=(5+12)^{9}=5\times (31\cdot 4481)^{2}&+12\times (3^{2}\cdot 19\cdot 251)^{2}\end{array}}$

Uniqueness of (x+y)ⁿ = xu²+yv²

Proposition

Let $p$ an odd prime and $(x,y)$ 2 positive integers such that $p=x+y$. Then there exists a unique pair of coprime positive integers $(u,v)$ such that $p^{n}=(x+y)^{n}=x\cdot u^{2}+y\cdot v^{2}$

Examples for prime and composite numbers:

The first form is always given by the square formula (2) .

The additional ones for composite numbers found with a python script

${\begin{aligned}3^{3}&=1\cdot 5^{2}&+&2\cdot 1^{2}\\5^{3}&=1\cdot 11^{2}&+&4\cdot 1^{2}\\&=3\cdot 3^{2}&+&2\cdot 7^{2}\\\\15^{3}&=1\cdot 41^{2}&+&14\cdot 11^{2}\\&=7\cdot 17^{2}&+&8\cdot 13^{2}\\&={\color {red}11}\cdot 1^{2}&+&{\color {red}4}\cdot 29^{2}={\color {red}11}\cdot 17^{2}&+&{\color {red}4}\cdot 7^{2}\\&={\color {green}13}\cdot 7^{2}&+&{\color {green}2}\cdot 37^{2}={\color {green}13}\cdot 1^{2}&+&{\color {green}2}\cdot 41^{2}\end{aligned}}$

Proof

here on math.stackexchange.com. I report here the Jandri's "elementary" and brilliant one:

Let $n$ an odd integer,  $p=a+b$ an odd prime and let  $(u,v)$ and $(u',v')$ be two pairs of coprime integers such that $p^{n}=au^{2}+bv^{2}=au'^{2}+bv'^{2}$.

Combining the equalities we obtain $p^{n}(v'^{2}-v^{2})=a(u^{2}v'^{2}-v^{2}u'^{2})$ then $p^{n}$ divides $(uv'-vu')(uv'+vu')$.

$p$ cannot divide simultaneously  $(uv'-vu')$ and $(uv'+vu')$  otherwise  divides $2uv'$ then  $u$ or $v'$ ; if  $p$ divides $u$ then $p$ divides $v$  : contradiction because  $u$ and $v$ are coprime (idem $p$ if  $v'$ divides ).

We deduce that $p^{n}$ divides $uv'-\varepsilon vu'$  with $\varepsilon =\pm 1$.

To finish we write by multiplying the two expressions of $p^{n}$ :

$p^{2n}=(auu'+\varepsilon bvv')^{2}+ab(uv'-\varepsilon vu')^{2}$ .

We deduce $uv'=\varepsilon vu'$  then $u=u'$ and $v=v'$.

Prime power: n-valuation and prime factors mod 2n

Here $n\in \mathbb {P}$

Let us rewrite propositions (6) and (7) in term of n-valuation:

Proposition

$n\in \mathbb {P} ,2\mid xy,~x\wedge y=1\Rightarrow {\begin{cases}n\nmid x\Rightarrow v_{n}(f_{n}(x,y))=0&(6)\\n\mid x\Rightarrow v_{n}(f_{n}(x,y))=1&(7)\end{cases}}$

Proofs

cf previously in coprimality

Proposition

$n\in \mathbb {P} ,2\mid xy,~x\wedge y=1,\quad {\begin{cases}f_{n}(x,y)=\prod p_{i}^{v_{i}}&\Rightarrow p_{i}\equiv \pm 1[2n]&(8)\\f_{n}(x^{2},y)=\prod p_{i}^{v_{i}}&\Rightarrow p_{i}\equiv \pm 1[2n]&(9)\\f_{n}(x^{2},y^{2})=\prod p_{i}^{v_{i}}&\Rightarrow p_{i}\equiv \ 1[2n]&(10)\end{cases}}$

Proofs

here for (8) p=1[2n] .

here for (10): math.stackexchange.com .Thanks Thomas Andrews

Note

Fermat theorem gives $f_{n}(x,y)\equiv \pm 1[n]$ and $f_{n}(x^{2},y)\equiv 1[n]$ . But it also applies to all the prime factors

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 ${\dfrac {2^{n}+1}{3}}$ 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 $f_{n}(x,y)\equiv \pm 1[2n]$

${\begin{array}{lll}f_{5}(6,11)&=1301&\equiv 1&[10]\\f_{7}(6,11)&=181\times 239&\equiv -1\times 1&[14]\\f_{11}(6,11)&=47820079&\equiv 1&[22]\\f_{13}(6,11)&=8969\times 177269&\equiv -1\times 1&[26]\\f_{19}(6,11)&=151\times 386989747169&\equiv -1\times -1&[38]\\f_{23}(6,11)&=5278223\times 12238317893&\equiv -1\times -1&[46]\\f_{29}(6,11)&=233\times 521\times 1309699\times 14932891583&\equiv 1\times -1\times 1\times -1&[58]\\...\end{array}}$

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

${\begin{array}{lll}f_{5}(6^{2},11)&=5861&\equiv 1&[10]\\f_{7}(6^{2},11)&=507809&\equiv 1&[14]\\f_{11}(6^{2},11)&=89\times 6029\times 7129&\equiv 1\times 1\times 1&[22]\\f_{13}(6^{2},11)&=883\times 2341\times 160627&\equiv -1\times 1\times -1&[26]\\f_{17}(6^{2},11)&=67\times 187067\times 199591969&\equiv -1\times -1\times 1&[34]\\f_{19}(6^{2},11)&=229\times 683\times 1901\times 2963\times 246469&\equiv 1\times -1\times 1\times -1\times 1&[38]\\f_{23}(6^{2},11)&=367\times 4457595074737380607&\equiv -1\times -1&[46]\\f_{29}(6^{2},11)&=1069838719517673460520580221&\equiv 1&[58]\\...\end{array}}$

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

${\begin{array}{lll}f_{5}(6^{2},11^{2})&=118061&\equiv 1&[10]\\f_{7}(6^{2},11^{2})&=34188379&\equiv 1&[14]\\f_{11}(6^{2},11^{2})&=23\times 89\times 2377\times 586961&\equiv 1\times 1\times 1\times 1&[22]\\f_{13}(6^{2},11^{2})&=30187\times 27342279823&\equiv 1\times 1&[26]\\f_{19}(6^{2},11^{2})&=2129\times 3079\times 3039225397425209&\equiv 1\times 1\times 1&[38]\\f_{23}(6^{2},11^{2})&=1663964075070633572800332299&\equiv 1&[46]\\f_{29}(6^{2},11^{2})&=59\times 11250493\times 60508154689065340703592763&\equiv 1\times 1\times 1&[58]\\...\end{array}}$