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We did in class $x^2+y^2$, which was easy, and we had for homework $2x^2+2xy+3y^2$, which I did (its values (if not square) must be divisible by form primes, or of the form $x^2+5y^2$, and clearly they can't all be the latter!).

Just to test my method I took the middle coefficient down by one - and ran into trouble...

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    Perhaps you ran into unexpected trouble because you picked a form of discriminant -23, which is the first time the class number is 3, so you'll need to juggle values of three quadratic forms.2010-12-25
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    Also, whereas in the other cases you looked at the primes that are represented by the given form are determined by a congruence condition modulo the discriminant (e.g. $p \equiv 1 \bmod 4$ in the case of $x^2 + y^2$, and conditions mod $20$ for the form $2 x^2 + 2 x y + 3 y^2$) there is no such congruence condition for determining the prime represented by your form of discriminant $23$.2010-12-25
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    +1 for trying to generalize an assignment-shows thinking about the issues.2010-12-26
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    Dear Simplequestions, Do the comments by me and KCd mean anything to you? If so, great! If not, is there some point you would like to have elaborated?2010-12-26
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    Thank you very much for the comments! Hopefully next semester I will learn a non-elementary proof :) (Hilbert Class Fields and L-Series - sounds like an interesting course)2010-12-26
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    You might find this book interesting: http://books.google.com/books?id=pSMlAQAAIAAJ2010-12-26
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    I'm missing something. Aren't there an infinite number of primes for y=1, ie "2*x^2*x+3". If you mod "2*x^2+2*x+3" by any prime p, there are at most two values of x (per p) that mod to 0. Can you build a proof from that?2011-03-13
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    @barrycarter: no. Actually a result of this form is not known for _any_ polynomial of degree greater than $1$.2011-03-29

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