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?
Thanks!
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).
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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 (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]
If we take the polynomial in terms of x, the discriminant of is . In order for x to be a real number, this value must be positive, and in order for x to be an integer, this value must be a square. The only integer values for y such that x is real are -1, 0, and 1, and none of these values makes the discriminant a square. Hence 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?
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?
Thank you!