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December 18th, 2016, 10:51 AM  #1 
Newbie Joined: Dec 2016 From: Saudi Arabia Posts: 5 Thanks: 0  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: (re+fq)(refq)=r^2(e^2+f^2)f^2(r^2+q^2)=pr^2nf^2 as p is prime so it's either divisor of (re+fq) or (refq). let's say p divides (re+fq), with this formula: (re+fq)^2 + (rfeq)^2=pn=p^2 (n/p) which gives (n/p)=(re+rq)^2/p^2 + (rfeq)^2/p^2 ! The other case is similar, but will give: (n/p)=(rq+re)^2/p^2 + (refq)^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: U0=0 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 = kq. *0<r<q (as kq>k/2>q) *k divides p^2+r^2 ( as p^2+r^2 = p^2+q^2 +k(k2q)) 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+(qdk)^2=p^2+q^2 +k(kd^22d) => 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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