• March 21st 2010, 03:01 AM
kingwinner
a) [solved] Show that $x^2 + 5y^2$ and $2x^2 + 2xy + 3y^2$ 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 $f(x,y) = ax^2 + bxy + cy^2$ 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 $x^2 + 5y^2$, then the Legendre symbol (p/5) =1 or 0, and that if p is a prime represented by the form $2x^2 + 2xy + 3y^2$, then (p/5)= -1.

d) By combining parts a & c, show that an odd prime p is represented by the form $x^2 + 5y^2$ if and only if p=5 or p≡1 or 9 (mod20), and that a prime p is represented by the form $2x^2 + 2xy + 3y^2$ 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 $(-1)^{(p - 1)/2}$(p/5)=1
Now $(-1)^{(p - 1)/2}$(p/5)=1 <=> $(-1)^{(p - 1)/2}$=(p/5)=1 or $(-1)^{(p - 1)/2}$=(p/5)= -1
$(-1)^{(p - 1)/2}$=(p/5)=1 <=> p≡1(mod 4) and p≡1,4(mod 5) <=>p≡1,9(mod 20)
$(-1)^{(p - 1)/2}$=(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 $x^2 + 5y^2$, then by part b, 4p≡ $z^2$ (mod 20) has a solution.
z must be even, so let z=2k
=> 4p≡ $4k^2$ (mod 20) => p≡ $k^2$ (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 $2x^2 + 2xy + 3y^2$, then by part b, 8p≡ $z^2$ (mod 20) has a solution.
z must be even, so let z=2k
=> 8p≡ $4k^2$ (mod 20) => 2p≡ $k^2$ (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?

Thank you! :)

[also under discussion in math links forum]
• March 22nd 2010, 02:37 PM
kingwinner
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 $2x^{2}+2xy+3y^{2} =5$ has no integer solutions?

Thanks!
• March 22nd 2010, 03:01 PM
icemanfan
Quote:

Originally Posted by kingwinner
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 $2x^{2}+2xy+3y^{2} =5$ has no integer solutions?

Thanks!

If we take the polynomial in terms of x, the discriminant of $2x^2 + 2xy + 3y^2 - 5 = 0$ is $20(2 - y^2)$. 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 $2x^2 + 2xy + 3y^2 = 5$ has no integer solutions.
• March 22nd 2010, 03:17 PM
chiph588@
Quote:

Originally Posted by kingwinner
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 $2x^{2}+2xy+3y^{2} =5$ has no integer solutions?

Thanks!

Consider the equation modulo $4$

$5 \equiv 1 \mod{4}$

There are $16$ pairings of $(x,y)$ consider for $f(x,y)=2x^2+2xy+3y^2$ modulo $4$.

It turns out they all come out to be either $0,2,$ or $3$.

So since we cant find $(x,y)$ such that $f(x,y) \equiv 1 \mod{4}$ we are done.
• March 22nd 2010, 04:35 PM
kingwinner
Quote:

Originally Posted by chiph588@
Consider the equation modulo $4$

$5 \equiv 1 \mod{4}$

There are $16$ pairings of $(x,y)$ consider for $f(x,y)=2x^2+2xy+3y^2$ modulo $4$.

It turns out they all come out to be either $0,2,$ or $3$.

So since we cant find $(x,y)$ such that $f(x,y) \equiv 1 \mod{4}$ we are done.

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!
• March 22nd 2010, 04:55 PM
chiph588@
Quote:

Originally Posted by kingwinner
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!

Well I started with mod 3 and when that didn't work, I just moved up the ladder. Luckily mod 4 worked.

And your reasoning is correct too.
• March 22nd 2010, 05:00 PM
kingwinner
I see.

Is there any alternative way to prove that $2x^{2}+2xy+3y^{2}=5$ has no integer solutions?

Say, by completing the square?
$
(x+y)^{2}+x^{2}+2y^{2} = 5
$

Is it possible to eliminate some cases from here and show the remaining cases do not give 5?
• March 23rd 2010, 05:11 PM
chiph588@
Quote:

Originally Posted by kingwinner
I see.

Is there any alternative way to prove that $2x^{2}+2xy+3y^{2}=5$ has no integer solutions?

Say, by completing the square?
$
(x+y)^{2}+x^{2}+2y^{2} = 5
$

Is it possible to eliminate some cases from here and show the remaining cases do not give 5?

Since all three terms are positive, $|y|\leq 1$ and $|x|\leq 2$.

You could probably find more constrains by looking at $x+y$, but the above two constraints gives us 15 cases to check, which is good enough for me (Rock).