Proof about a diophantine equation

[cragar]

Homework Statement

Prove that the Diophantine equation
$3y^2-1=x^2$
has no solution.

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 don't know if that's a good place to start.

 

[Dick]

Homework Statement

Prove that the Diophantine equation
$3y^2-1=x^2$
has no solution.

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 don't know if that's a good place to start.

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?

 

[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.

 

[Dick]

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.

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

 

[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 won't work.

 

[Dick]

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 won't work.

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?

 

[CAF123]

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

 

[Dick]

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

You could also do that. Sure.