a) [solved] Show that $\displaystyle x^2 + 5y^2$ and $\displaystyle 2x^2 + 2xy + 3y^2$ are the only reduced positive definite binary quadratic forms of discriminant -20. Show that the first of these forms does not represent 2, but that the second one does, so that these forms are inequivalent, and hence that the class number H(-20)=2. Show that a prime p is represented by at least one of these forms if and only if p=2, p=5, or p≡1,3,7, or 9 (mod20). [solved]

b) [solved] Show that if a number n is represented by a binary quadratic form $\displaystyle f(x,y) = ax^2 + bxy + cy^2$ of discriminant d, then 4an is a square modulo |d|. [solved]

c) Using part b, show that if a prime p is represented by the form $\displaystyle x^2 + 5y^2$, then the Legendre symbol (p/5) =1 or 0, and that if p is a prime represented by the form $\displaystyle 2x^2 + 2xy + 3y^2$, then (p/5)= -1.

d) By combining parts a & c, show that an odd prime p is represented by the form $\displaystyle x^2 + 5y^2$ if and only if p=5 or p≡1 or 9 (mod20), and that a prime p is represented by the form $\displaystyle 2x^2 + 2xy + 3y^2$ if and only if p=2 or p≡3 or 7 (mod 20).

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I proved parts a & b, but I'm stuck on parts c & d.

a) An odd prime p is represented by at least one of these forms of discriminant -20

<=> p|(-20) or (-20/p)=1

<=> p=5 or (-20/p)=1

<=> p=5 or $\displaystyle (-1)^{(p - 1)/2}$(p/5)=1

Now $\displaystyle (-1)^{(p - 1)/2}$(p/5)=1 <=> $\displaystyle (-1)^{(p - 1)/2}$=(p/5)=1 or $\displaystyle (-1)^{(p - 1)/2}$=(p/5)= -1

$\displaystyle (-1)^{(p - 1)/2}$=(p/5)=1 <=> p≡1(mod 4) and p≡1,4(mod 5) <=>p≡1,9(mod 20)

$\displaystyle (-1)^{(p - 1)/2}$=(p/5)= -1 <=> p≡3(mod 4) and p≡2,3(mod 5) <=>p≡3,7(mod 20)

c) If p is a prime represented by $\displaystyle x^2 + 5y^2$, then by part b, 4p≡$\displaystyle z^2$ (mod 20) has a solution.

z must be even, so let z=2k

=> 4p≡$\displaystyle 4k^2$ (mod 20) => p≡$\displaystyle k^2$ (mod 5)

=> 5|p or (p/5)=1 => p=5 or (p/5)=1 => (p/5)=0 or 1.

If p is a prime represented by $\displaystyle 2x^2 + 2xy + 3y^2$, then by part b, 8p≡$\displaystyle z^2$ (mod 20) has a solution.

z must be even, so let z=2k

=> 8p≡$\displaystyle 4k^2$ (mod 20) => 2p≡$\displaystyle k^2$ (mod 5)

=> 5|2p or (2p/5)=1 => 5|p or (2p/5)=1

=> p=5 or (2p/5)=1

=> p=5 or (p/5)= -1

=> (p/5)= 0 or -1 which is NOT what we want, but I can't find where my mistake is...

d) I'm not sure how to "combine" the information in parts a & c. In particular, we are required to prove "if and only if" in part d. How can we prove it?

Can someone help me, please?

Thank you! :)

[also under discussion in math links forum]