Talk:Bézout's identity

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I should do the searching... but just in case. Is it true that one can get the coefficients of less degree than the starting elements if one is in an Euclidean Domain? In the case of integers this amounts to saying that

there are x,y ∈ Z wiht |x|<max(a,b), |y|<max(a,b) with xa+yb=gcd(a,b)

Just in case. I'll try to find out though :) Pfortuny 09:26, 20 Apr 2004 (UTC)

[edit] proof

well, any proof is a proof (if it is one), but I like quite well the one on the french version of this page (considering the smallest element of { ax+by\in\N^* } which divides a and b.) — MFH:Talk 19:13, 6 October 2006 (UTC)

That being said, I am worried about this sentence from the first (English) proof:

Now, consider the numbers p, 2p, …, (q−1)p. None of these numbers is congruent to 0 modulo q, and they are also all distinct modulo q.

The statement is unassailable, but every proof of it I know uses either this identity, or prime factorisation, which, again, is almost invariably proven using this identity. Can it be justified independently? JadeNB 22:20, 8 March 2007 (UTC)

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