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December 18th, 2016, 10:51 AM   #1
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Lightbulb divisors of sum of two squares

Hi Again,

In this thread, I am attempting a proof for the following theorem:

every divisor of two squares of coprimes is a sum of two squares.

1. For this, I will start by proving the following: (C1)
if n is a sum of two squares and p is a prime divisor who is sum of two squares than n/p is also a sum of two squares.

In all the following, we will use:

n = r^2 + q^2 and p=e^2 + f^2. (r,q coprime)

than we have this formula:

as p is prime so it's either divisor of (re+fq) or (re-fq).
let's say p divides (re+fq), with this formula:

(re+fq)^2 + (rf-eq)^2=pn=p^2 (n/p)

which gives (n/p)=(re+rq)^2/p^2 + (rf-eq)^2/p^2 !
The other case is similar, but will give:

(n/p)=(rq+re)^2/p^2 + (re-fq)^2/p^2 !

2.Now we will attack the theorem using T0.
First, some definitions:

let P be the set of all integers that are sum of squares of two coprimes.

P={p =u^2+v^2, u and v integers coprime}

let (Un) be a sequence on P, defined by:

Ui+1 = min{p E P, p>Ui} (I am using here E for "belongs to")

->Properties that are a consequence of the definitions:
*every element p of P has an i such as Ui=p.
*Un is increasing.
*U1=1, U2=2, U3=5, U4=10.

Clearly, T1 is true for Up for p<5.

With this, for a given n>0, let's suppose T1 true for all Uk, with<n, and let's prove it's true for Un.
we'll take Un=p^2 + q^2 (q>=p)

1. if Un is prime than it's OK.
2. When not, let k be a divisor of Un that is less than p+q. This exists because p^2+q^2< (p+q)^2.

so we have naturally q>k/2.

We have two cases:
2.1 q<k
let r = k-q.
*0<r<q (as k-q>k/2>q)

*k divides p^2+r^2 ( as p^2+r^2 = p^2+q^2 +k(k-2q))
2.2 q>k
let r be the rest of the Euclidean division of q by k, and d the divisor.

by definition of the Euclidean division r<q.

p^2+r^2=p^2+(q-dk)^2=p^2+q^2 +k(kd^2-2d) => k divise p^2 + r^2

so in all cases k is also a divisor of p^2 + r^2, or by our induction supposition every divisor of p^2 + r^2 is sum of two squares, which gives k is a prime divisor of two squares.

And Using T0 we have (Un/k) is also a sum of two squares, and using the supposition again Un/k have all its divisors sum of two squares.

which means Un have all its divisors as sum of two squares! (*)

(*) I supposed here known that product of two elements of P is also an element of P.

That's it ! I hope you will find it somewhat consistent.

Last edited by skipjack; December 18th, 2016 at 11:47 AM.
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