Is the equation 4x^(2) + 5y^(2) = 2 solvable with rational roots?

[saadsarfraz]

Show that the equation 4x^(2) + 5y^(2) = 2 has no rational solutions.

Can this be done graphically?

 

[Hurkyl]

If you have a rigorous graphical method, then of course. Though, I'm hard pressed to think of any that would be applicable...

 

[MathematicalPhysicist]

Suppose the equation has rational roots:
4x^(2) + 5y^(2) = 2
x=p1/q1
y=p2/q2
gcd(p1,q1)=gcd(p2,q2)=1
4(p1^2/q1^2)+5(p2/q2)^2=2
[4(p1q2)^2+5(p2q1)^2]/(q2q1)^2=2
4(p1q2)^2+5(p2q1)^2=2(q2q1)^2
then the question becomes does the equation:
4a+5b=2c
where a,b,c are natural numbers has solution.
4(a+b)+b=2c
because 2c is even b must be even also, but then b=(4a-2c)/5 then 4a-2c is divisble by 5, i.e 4a-2c=0 or a and c are multiples of 5.
if b=0 then c=2a i.e (q2q1)^2=2(p1q2)^2 but this isn't possible cause sqrt(2) is irrational.
Now you need to check the option of multiples of 5 and get a contradiction, perhaps someone else with more intel in number theory will chip in with this question.