The quadratic form $x^2+ny^2$ via prime factors - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T15:14:26Z http://mathoverflow.net/feeds/question/111489 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111489/the-quadratic-form-x2ny2-via-prime-factors The quadratic form $x^2+ny^2$ via prime factors Stemkoski 2012-11-04T22:00:15Z 2012-11-05T03:53:41Z <p>Elementary algebra shows that the product of two numbers in the form $x^2 + ny^2$ again has the same form, since if $p = (a^2 + nb^2)$ and $q = (c^2 + nd^2)$, $$pq = (a^2 + nb^2)(c^2 + nd^2) = (ac \pm nbd)^2 + n(ad \mp bc)^2$$ My question is: Assuming that a number $z$ can be factored into primes of the form $x^2 + ny^2$, does <em>every</em> representation of $z$ in this form arise from repeated applications of this formula to the prime factors? </p> http://mathoverflow.net/questions/111489/the-quadratic-form-x2ny2-via-prime-factors/111502#111502 Answer by GH for The quadratic form $x^2+ny^2$ via prime factors GH 2012-11-05T01:30:30Z 2012-11-05T01:56:25Z <p>The answer is yes. To see this, consider the ring $R=\mathbb{Z}[\sqrt{-n}]$. If $z=p_1\dots p_k$ is the decomposition of $z$ into rational primes, then by assumption each $p_j$ decomposes in $R$ as $p_j=q_j\bar q_j$. We need to show that any decomposition $z=r\bar r$ in $R$ can be gotten as follows: for each $j$ let $r_j$ be either $q_j$ or $\bar q_j$, and then put $q=wr_1\dots r_k$, where $w$ is a unit in $R$. In $R$ the ideal $(z)$ decomposes into prime ideals as $(z)=(q_1)(\bar q_1)\dots (q_k)(\bar q_k)$, hence it suffices to show that in $R$ the ideal $(z)$ and its divisors decompose uniquely into prime ideals.</p> <p>If $n$ is square-free and congruent to $1$ or $2$ mod $4$, then $R$ is the full ring of integers in $\mathbb{Q}(\sqrt{-n})$, hence it is a Dedekind domain. So in this case we are done.</p> <p>If $n$ is square-free and congruent to $3$ mod $4$, then $R$ is a quadratic order of conductor $2$ in $\mathbb{Q}(\sqrt{-n})$, hence unique factorization holds in $R$ for ideals prime to $2$. Clearly, each $p_j$ above is odd, hence $(z)$ is coprime to $2$, and we are done.</p> <p>If $n$ is not square-free, then $R$ is a quadratic order of some conductor $f\mid 2n$ in $\mathbb{Q}(\sqrt{-n})$, hence unique factorization holds in $R$ for ideals prime to $f$. Clearly, each $p_j$ above is at least $n$, hence $(z)$ is coprime to $2n$, and we are done.</p> <p>For the quoted result on quadratic orders see Exercise 7.26 in Cox: Primes of the form $x^2+ny^2$.</p> http://mathoverflow.net/questions/111489/the-quadratic-form-x2ny2-via-prime-factors/111518#111518 Answer by Will Jagy for The quadratic form $x^2+ny^2$ via prime factors Will Jagy 2012-11-05T03:53:41Z 2012-11-05T03:53:41Z <p>Why not. My answer at <a href="http://math.stackexchange.com/questions/229201/the-quadratic-form-x2-ny2-via-prime-factors/229270#229270" rel="nofollow">http://math.stackexchange.com/questions/229201/the-quadratic-form-x2-ny2-via-prime-factors/229270#229270</a> </p>