Talk:Proofs of Fermat's theorem on sums of two squares
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Two more proofs can be seen here: http://planetmath.org/encyclopedia/ProofOfThuesLemma.html (I don't know why they call the result Thue's Lemma.)
- Feel free to add an outline of those proofs, or to add the reference. Magidin 15:08, 21 February 2006 (UTC)
I added a link to those other proofs. Is that you, Arturo? :) LDH 10:54, 22 February 2006 (UTC)
[edit] Don Zagier's proof
A recent addition to Don Zagier gives references to a short proof that perhaps deserves mention here: [1][2]. 165.189.91.148 20:43, 14 September 2006 (UTC)
[edit] Unique Factorisation Domain
It seems misleading to say "Since Z[i] is a unique factorization domain (in fact, a Euclidean domain), every ideal is principal."
Z[i] is a Euclidean domain and as such it is a PID, but this is not guaranteed by the fact that it is a unique factorisation domain.
Suggest the following revision: "Since Z[i] is a principal ideal domain (in fact, a Euclidean domain), every ideal is principal." Or, alternatively, just say that this follows from the fact that Z[i] is a Euclidean domain. —The preceding unsigned comment was added by 163.1.146.236 (talk) 18:27, 1 February 2007 (UTC).
- I made the change, but I wanted to point out the reason it was written as it was: basically, it was paraphrased from Dedekind's argument. For rings of integers of number fields, the ccondition of being a PID and being a UFD are equivalent; at the time, these were the only general "rings" of interest. Dedekind was interested n unique factorization, not the PID property, though he had shown that the two were equivalent. Magidin 19:57, 3 February 2007 (UTC)
[edit] Euler's proof
In Euler's proof, in his second statement, it is not necessary to state that p2 +q2 divides (ap+bq)2, because in the final expression, a fraction is left out. But a better statement would be that the lift hand side of the final expression is an integer and the second term in the right hand side is an integer, so the first term better be an integer. This better be because, it does not mean that (p2 + q2)2 divides (ap+bq)2 from the above statement
- Whether this is better or not is not really material. The proof is as it was quoted in Edwards's book, and is meant to be the proof that was actually given by Euler. No doubt it can be improved, but that is not the point here. Also, please remember to sign and date your comments. Magidin 18:24, 25 March 2007 (UTC)
The first term in the left hand side of the expression is ((ap+bq)÷(p2 + q2))2. So (p2 + q2)2 ÷ (ap+bq)2 is an integer. Therefore the remaining is a fraction (1÷(p2 + q2)). This cannot be because the quotient is an integer and the second term in the right hand side is also an integer, therefore the first term better be an integer. This is contradictory.
