wu :: forums
« wu :: forums - rephrased: Simpler proof for Thue's lemma - ? »

Welcome, Guest. Please Login or Register.
May 15th, 2024, 9:53pm

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   riddles
   easy
(Moderators: Grimbal, Eigenray, ThudnBlunder, towr, Icarus, william wu, SMQ)
   rephrased: Simpler proof for Thue's lemma - ?
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: rephrased: Simpler proof for Thue's lemma - ?  (Read 1846 times)
Mickey1
Junior Member
**





   


Gender: male
Posts: 116
rephrased: Simpler proof for Thue's lemma - ?  
« on: Sep 26th, 2010, 7:38am »
Quote Quote Modify Modify

One often see Fermat’s sum of two squares theorem but less often the uniqueness of the representation is mentioned. The riddle here is to check that the proof is OK. I wonder if it is correct because it seems much simpler than Thue's lemma mentioned elsewhere  (search the forum for "Fermat"). Also when I search elsewhere (google), Thue's lemma seems to cover a lot of other things, so I am grateful for any help.  
 
My approach: Assume there were more than one way of representing primes:  
p (= 4n+1) = x1^2+y1^2= x2^2+y2^2. One of the squares most be even and the other odd, since p is odd.  Let x1^2, and therefore x1, be even and let y2 be odd (giving a simpler proof - see the example for 3*5 as he generating product, 4*6 would be much more complicated).
 
Rearrange to get x1^2 – y2^2= x2^2 – y1^2 so that (x1+y2)(x1-y2)=( x2+y1)( x2-y1).  x1+y2 and x1-y2 must also be odd, x1 being even and y2 being odd, so that their product (x1+y2)(x1-y2) is also odd.
 
Rename these two factors a and b, which are both odd.  Now we rewrite the product ab in two ways,      a * b and (ab)*1 .
 
The first factor a is written ((a+b)/2 + (a-b)/2)  
the second factor b is written ((a+b)/2 - (a-b)/2)  
 
so that  
a*b = ((a+b)/2 + (a-b)/2) ((a+b)/2 - (a-b)/2) = ((a+b)/2)^2 – ((a-b)/2)^2     (1)
 
The same procedure is done for (ab)*1:
(ab)*1 = (ab+1)/2 + (ab-1)/2) ((ab+1)/2 - (ab-1)/2) = ((ab+1)/2)^2 - ((ab-1)/2)^2  (2)
ab = (1)= (2) gives us  
   ((ab+1)/2)^2 - ((ab-1)/2)^2 = ((ab+1)/2)^2 - ((ab-1)/2)^2  which we rearrange to get the two sums of squares back = p
p = ((a^2+1)/2 )*((b^2+1)/2)               (3)  
As an example 5*3 can be written  
3*5 = (4-1)(4+1)= 16-1=4^2-1^2 and (5*3)*1 = (8+7)(8-7)=8^2-7^2  
And the corresponding two sums of squares are 8^2+1=7^2+4^2 = 65  
65 is also (1/4)(5*5+1)(3*3+1) according to formula (3). The two differences can only be arranged to yield the two sums of two squares in one way.  
« Last Edit: Nov 18th, 2010, 3:00pm by Mickey1 » IP Logged
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board