1. Feb 8, 2013

cragar

1. The problem statement, all variables and given/known data
Prove that the Diophantine equation
$3y^2-1=x^2$
has no solution.
3. The attempt at a solution
If we factor the left side as
$(\sqrt{3}y+1)(\sqrt{3}y-1)$
I was going to try to argue that x could only divide one of those.
But I dont know if thats a good place to start.

2. Feb 9, 2013

Dick

No. It's a really awful place to start. Diophantine means you are supposed to get solutions that are integers. You know that, don't you? What other ways have you seen to prove stuff like this?

Last edited: Feb 9, 2013
3. Feb 9, 2013

cragar

I know they need to be integers, I was trying to think of a way to argue that they had different amount of prime divisors. Or maybe it has something to to with their gcd's.

4. Feb 9, 2013

Dick

Those are more reasonable things to think about. How about trying to see if you can solve the equation modulo some number?

5. Feb 9, 2013

cragar

What about this if x is even then $3y^2$must be odd
so we have $3y^2=2a+1$ and then we
have $3y^2-1=2a=x^2$
if x=2 then this would work. but if a has more factors than just 2.
if a is a perfect square then it will either contain a factor of 2 or not in either case
it will have an odd number of factors of 2 but x has an even number of factors.
So this wont work.

6. Feb 9, 2013

Dick

Sure. If a is a perfect square then 2a=x^2 can't be solved. That's basically how to prove sqrt(2) is irrational. But why would you think a needs to be a perfect square? Here's a big hint. If a is any integer, what can you say about a^2 modulo 4?

7. Feb 11, 2013

CAF123

Can we not just consider the equation $x^2 \equiv -1$mod $3$, and show from here that this has no solutions?

8. Feb 11, 2013

Dick

You could also do that. Sure.