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

Can this be done graphically?

Can this be done graphically?

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- Thread starter saadsarfraz
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- #1

- 86

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

Can this be done graphically?

Can this be done graphically?

- #2

Hurkyl

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MathematicalPhysicist

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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.

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