Binomial theorem and odd power

This paper deals with a remarkable square form $\mathbf{(}𝐱\mathbf{+}𝐲{\mathbf{)}}^{𝐧}\mathbf{=}𝐱\mathbf{\cdot }{𝐜}^{\mathrm{𝟐}}\mathbf{+}𝐲\mathbf{\cdot }{𝐝}^{\mathrm{𝟐}}$ where $𝐧$ is an odd positive integer. Its existence comes from the symetrical grouping of binomial terms into a form $(x+y{)}^n=x\cdot a+y\cdot b$, First we give the algebric expressions of these $(a,b,c,d)$ coefficients. Then we focus on some properties in $\mathbb{Z}$ . With $(x,y)$ 2 coprime integers with different parity, we show that $(a,b)=1$ and $(c,d)=1$. We also show that there is uniqueness of $(c,d)$ when $x+y$ is prime . Finally, considering exponent $n$ prime, we show that the prime factors of $(a,b)$ are congruent to $1[2n]$, whereas $(c,d)$ are 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

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

Example

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

Algebraic properties

Propositions

$\begin{array}{ccccc}(x-y{)}^n& =x{f}_n(x^2,y^2)& -& y{f}_n(y^2,x^2)& (1)\\ \mathbf{(}𝐱\mathbf{-}𝐲{\mathbf{)}}^{𝐧}& =𝐱{𝐟}_{𝐧}\mathbf{(}𝐱,𝐲{\mathbf{)}}^{\mathrm{𝟐}}& -& 𝐲{𝐟}_{𝐧}\mathbf{(}𝐲,𝐱{\mathbf{)}}^{\mathrm{𝟐}}& (2)\end{array}$

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

$\begin{array}{ccc}(x+y{)}^n+(x-y{)}^n& =2x{f}_n(x^2,y^2)& (3)\\ x^n+y^y& =\frac{1}{2^{n-1}}(x+y)f_n((x+y{)}^2,(x-y{)}^2)& (3bis)\end{array}$

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 $f_n(x,y)$ :

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

Proposition

$2\mid xy,x\wedge y=1\Rightarrow \{\begin{array}{cc}{f}_n(x,y)\wedge f_n(y,x)=1& (5)\\ n\nmid x\Rightarrow x\wedge f_n(x,y)=1& (6)\\ n\mid x\Rightarrow x\wedge f_n(x,y)=n& (7)\end{array}$

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 $y$

Examples: previous examples with prime factors

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

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

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

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 $(u^{\prime },v^{\prime })$ be two pairs of coprime integers such that $p^n=a{u}^2+b{v}^2=au{\text{'}}^2+bv{\text{'}}^2$.

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

$p$ cannot divide simultaneously  $(u{v}^{\prime }-v{u}^{\prime })$ and $(u{v}^{\prime }+v{u}^{\prime })$  otherwise  divides $2u{v}^{\prime }$ then  $u$ or $v^{\prime }$ ; if  $p$ divides $u$ then $p$ divides $v$  : contradiction because  $u$ and $v$ are coprime (idem $p$ if  $v^{\prime }$ divides ).

We deduce that $p^n$ divides $u{v}^{\prime }-\epsilon v{u}^{\prime }$  with $\epsilon =\pm 1$.

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

$p^{2n}=(au{u}^{\prime }+\epsilon bv{v}^{\prime }{)}^2+ab(u{v}^{\prime }-\epsilon v{u}^{\prime }{)}^2$ .

We deduce $u{v}^{\prime }=\epsilon v{u}^{\prime }$  then $u=u^{\prime }$ and $v=v^{\prime }$.

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

$n\in \mathbb{P},2\mid xy,x\wedge y=1\Rightarrow \{\begin{array}{cc}n\nmid x\Rightarrow v_n(f_n(x,y))=0& (6)\\ n\mid x\Rightarrow v_n(f_n(x,y))=1& (7)\end{array}$

Proofs

cf previously in coprimality

Proposition

$n\in \mathbb{P},2\mid xy,x\wedge y=1,\{\begin{array}{ccc}{f}_n(x,y)=\prod p_i^{v_i}& \Rightarrow p_i\equiv \pm 1[2n]& (8)\\ f_n(x^2,y)=\prod p_i^{v_i}& \Rightarrow p_i\equiv \pm 1[2n]& (9)\\ f_n(x^2,y^2)=\prod p_i^{v_i}& \Rightarrow p_i\equiv 1[2n]& (10)\end{array}$

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

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

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

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

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

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