# Thread: Solving X^3 = Y^2 + 5

1. ## Solving X^3 = Y^2 + 5

Hi,

I thought I had a reasonable understanding of Diophatine Equations (or atleast I thought I was fairly competent in solving them) until I saw this example that had been worked through in my lecture notes. I would normally email my lecturer if I had a problem though this one isn't particularly good at replying! Any help would be greatly appreciated as I have an exam on the topic looming.

Anyway, the problem is as follows: find the integer solutions to the equation X^3 = Y^2 + 5.

The solution first noted that Y^2 + 5 = (Y + (-5)^(1/2))(Y - (-5)^(1/2)) and so the highest common factor of (Y + (-5)^(1/2)), (Y - (-5)^(1/2)) will devide the difference 2(-5)^(1/2). Now from my understanding of the topic this would give rise to us considering the behaviour of the prime 2 in Q[(-5)^(1/2)] however, the solution considered also the prime 5. Why is this so?

Thanks for any help...

2. Originally Posted by TheFinalPush
Hi,

I thought I had a reasonable understanding of Diophatine Equations (or atleast I thought I was fairly competent in solving them) until I saw this example that had been worked through in my lecture notes. I would normally email my lecturer if I had a problem though this one isn't particularly good at replying! Any help would be greatly appreciated as I have an exam on the topic looming.

Anyway, the problem is as follows: find the integer solutions to the equation X^3 = Y^2 + 5.

The solution first noted that Y^2 + 5 = (Y + (-5)^(1/2))(Y - (-5)^(1/2)) and so the highest common factor of (Y + (-5)^(1/2)), (Y - (-5)^(1/2)) will devide the difference 2(-5)^(1/2). Now from my understanding of the topic this would give rise to us considering the behaviour of the prime 2 in Q[(-5)^(1/2)] however, the solution considered also the prime 5. Why is this so?

Thanks for any help...
If you have a solution (whether you understand it or not), could you post it? This would make things go a lot quicker.

3. I have found the solution but am having trouble posting it up, mathematical notation in tact. Until I can work out how (or someone tells me) here is the link to it. It is question 2.

http://www.maths.manchester.ac.uk/un...swersheet9.pdf

I think there are two points I need clarifying:

How can we tell from the h.c.f. which primes we need to consider?

If the class number is 2 why does class(B) = class(B^3)? Why does this mean the B is principal? I think these two questions probably result in my limited understanding of the quotient group Ck.

Thanks for any help, it is greatly appreciated...

4. Originally Posted by TheFinalPush
I have found the solution but am having trouble posting it up, mathematical notation in tact. Until I can work out how (or someone tells me) here is the link to it. It is question 2.

http://www.maths.manchester.ac.uk/un...swersheet9.pdf

I think there are two points I need clarifying:

How can we tell from the h.c.f. which primes we need to consider?

If the class number is 2 why does class(B) = class(B^3)? Why does this mean the B is principal? I think these two questions probably result in my limited understanding of the quotient group Ck.

Thanks for any help, it is greatly appreciated...
Nice proof, it looks similar to the proof I was working on for this problem, so that's encouraging!

Anyway, let $\displaystyle d=(y+\sqrt{-5},y-\sqrt{5})$. Perhaps it's because $\displaystyle d\mid2\sqrt{-5}\implies d\mid(2\sqrt{-5})^2=-4\cdot5$.

As for the class number, I'm not sure. This is where I hit a snag in my proof... Unfortunately I'm not familiar with class numbers yet.

5. Originally Posted by TheFinalPush
I have found the solution but am having trouble posting it up, mathematical notation in tact. Until I can work out how (or someone tells me) here is the link to it. It is question 2.

http://www.maths.manchester.ac.uk/un...swersheet9.pdf

I think there are two points I need clarifying:

How can we tell from the h.c.f. which primes we need to consider?

If the class number is 2 why does class(B) = class(B^3)? Why does this mean the B is principal? I think these two questions probably result in my limited understanding of the quotient group Ck.

That the class number = 2 means that every fractional ideal (remember?) squared is principal ( hint: if G is a group and K is a normal sbgp. of G of index n, then $\displaystyle g^n\in K\,,\,\,\forall\,g\in G$ ).

In our case , $\displaystyle B=B^3\iff B^2=1$ , which is true precisely because the class number is 2...and yes: perhaps you need to dive in a little deeper into that quotient group.

Tonio

Thanks for any help, it is greatly appreciated...
.

6. Originally Posted by tonio
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Tonio,

Would you happen to know of any good online references to learn about class numbers?

7. Originally Posted by chiph588@
Tonio,

Would you happen to know of any good online references to learn about class numbers?

Not really...perhaps googling "class number" and reading the references in Wikipedia, Wolfram and etc.
But there are several excellent books, mostly in algebraic number theory, that deal with this beautiful theme. This requires a good knowledge of basic group theory, ring theory and some module theory, which I think won't be a big problem for you.

Let me know if you're interested.

Tonio