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Dec 11, 2017 at 20:24 history closed Lucia
Pace Nielsen
Henry.L
Alex B.
Greg Martin
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Dec 11, 2017 at 20:15 comment added square-free @nfdc23, It is not a homework problem or take-home exam question.
Dec 11, 2017 at 20:13 vote accept square-free
Dec 11, 2017 at 16:25 history edited GH from MO
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Dec 11, 2017 at 16:11 answer added GH from MO timeline score: 8
Dec 11, 2017 at 15:38 review Close votes
Dec 11, 2017 at 20:27
Dec 11, 2017 at 13:02 comment added nfdc23 Is this a homework problem or take-home exam question? What is the motivation for asking about it here? There is such an $n$ that works, and this is a standard exercise in the initial parts of algebraic number theory. As a hint, there is an $n$ which even works for all $p\equiv 1 \bmod 3$.
Dec 11, 2017 at 12:20 comment added square-free @TobiasKildetoft, Yes I would like, that $n$ be different from $-1$
Dec 11, 2017 at 12:19 history edited Joe Silverman CC BY-SA 3.0
Fixed grammar
Dec 11, 2017 at 12:19 comment added square-free Prof @JoeSilverman, I don't have access to this book.
Dec 11, 2017 at 12:17 comment added Joe Silverman Have you read David Cox's book *Primes of the form $X^2+nY^2$"?
Dec 11, 2017 at 12:15 comment added Tobias Kildetoft So presumably you also want to specify that $n$ should be positive, as otherwise $n=-1$ obviously works. Once that is done my guess would be that there is no such $n$.
Dec 11, 2017 at 12:11 history edited square-free CC BY-SA 3.0
added 29 characters in body
Dec 11, 2017 at 12:08 comment added Gerry Myerson Then please edit your question to say so.
Dec 11, 2017 at 12:07 comment added square-free @GerryMyerson, Exactly, I would like to find an integer $n$ which doesn't depend on $p$.
Dec 11, 2017 at 12:07 comment added Gerry Myerson But $x=y=1$ satisfies $(x,y)=1$. You'll have to do better than that!
Dec 11, 2017 at 12:06 comment added square-free @JasonStarr, Thanks for your comment, I am trying to find a representation of $p$ in the form $x^2+ny^2$ with $(x,y)=1$.
Dec 11, 2017 at 12:06 comment added Gerry Myerson I imagine, Jason, that OP doesn't want $n$ to depend on $p$. Perhaps OP could edit the question, to clarify this point?
Dec 11, 2017 at 12:04 comment added Jason Starr Set $n$ equal to $p-1$ and set $x$ and $y$ equal to $1$. Is that what you are asking?
Dec 11, 2017 at 12:01 history asked square-free CC BY-SA 3.0