Binary Quadratic Forms
a) [solved] Show that and 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 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 , then the Legendre symbol (p/5) =1 or 0, and that if p is a prime represented by the form , then (p/5)= -1.
d) By combining parts a & c, show that an odd prime p is represented by the form if and only if p=5 or p≡1 or 9 (mod20), and that a prime p is represented by the form if and only if p=2 or p≡3 or 7 (mod 20).
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 (p/5)=1
Now (p/5)=1 <=> =(p/5)=1 or =(p/5)= -1
=(p/5)=1 <=> p≡1(mod 4) and p≡1,4(mod 5) <=>p≡1,9(mod 20)
=(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 , then by part b, 4p≡ (mod 20) has a solution.
z must be even, so let z=2k
=> 4p≡ (mod 20) => p≡ (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 , then by part b, 8p≡ (mod 20) has a solution.
z must be even, so let z=2k
=> 8p≡ (mod 20) => 2p≡ (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]
So for part c, I think we have to show that the case p=5 is impossible.
In short, my question just translates to the following:
How can we prove that has no integer solutions?
Why did you choose mod 4 (rather than mod something else)? How did you get this inspiration to work with mod 4? It's kind of mysterious to me, and I have no idea how to come up with inspirations like this...Can you explain the thinking behind it?
Originally Posted by chiph588@
Also, the logic of your proof is the following, right?
The conrguence 2x^2 + 2xy +3y^2 ≡ 5 (mod 4) has no solutions, thus the equation 2x^2 + 2xy +3y^2 = 5 has no integer solutions, right?
Well I started with mod 3 and when that didn't work, I just moved up the ladder. Luckily mod 4 worked.
Originally Posted by kingwinner
And your reasoning is correct too.
Is there any alternative way to prove that has no integer solutions?
Say, by completing the square?
Is it possible to eliminate some cases from here and show the remaining cases do not give 5?