An elementary proof of Gauss' genus theorem
Let \(m\neq 1\) be a square-free integer and \(d\) the discriminant of the quadratic field \(K=\mathbb{Q}(\sqrt m)\). Let \(p_ 1,\dots,p_ s\) be the odd prime divisors of \(n\). If \(A\) and \(B\) are ideals in \(K\) such that \(A=(\rho)B\) and \(N_ \rho>0\) we write \(A\sim B\). Gauss' duplication theorem asserts that if \(A\) is an ideal coprime with \(d\) then \(A\sim B^ 2\) for some ideal \(B\) if and only if \(\left({NA\over p_ i}\right)=1\), \(1\leq i\leq s\). The authors give a beautiful elementary proof of this famous result, which makes use of Legendre's familiar necessary and sufficient condition for the equation \(ax^ 2+by^ 2=z^ 2\) to have a non-trivial integral solution. The proof only takes about twenty lines and one's reaction is ``Now why didn't I think of that?'' (The ideal \(A\) should be nonzero in Proposition 1).
scientific article; zbMATH DE number 97937