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Topic: rephrased: Simpler proof for Thue's lemma - ? (Read 1846 times) |
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Mickey1
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rephrased: Simpler proof for Thue's lemma - ?
« on: Sep 26th, 2010, 7:38am » |
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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.
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« Last Edit: Nov 18th, 2010, 3:00pm by Mickey1 » |
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