Can we parametriza $x$ and $y$ in rational field $Q$ so that $$-8xy+8x^2-4y^2$$ be square?
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It is necessary to solve such an equation? $8x^2-8xy-4y^2=z^2$ – individ Jun 12 '14 at 08:51
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If so, you have to use this formula. http://math.stackexchange.com/questions/738446/solutions-to-ax2-by2-cz2/738527#738527 For ease of calculation will reduce by 4. Then we will use the equivalent of a quadratic form by replacing $x$ on $x+y$ – individ Jun 12 '14 at 09:12
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Oh sign't noticed. In integers decisions no. – individ Jun 12 '14 at 09:44
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If you can find one rational point on the curve, then you can take the line through that point with slope $t$, and the other interswection of line and curve will be a rational point, parametrized by $t$. – Gerry Myerson Jun 21 '14 at 13:51
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Any questions about the answer I have posted? – Gerry Myerson Jun 25 '14 at 11:50
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Are you still there? – Gerry Myerson Jun 26 '14 at 13:03
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The requested parametrization cannot exist, since the equation has no rational solutions.
There are nonzero rational $x,y$ such that $-8xy+8x^2=4y^2$ is a (rational) square if and only if there are nonzero integers $x,y,z$ such that $$2x^2-2xy-y^2=z^2$$ from which we get $$4x^2-4xy-2y^2=2z^2=(2x-y)^2-3y^2$$ and $$(2x-y)^2\equiv2z^2\pmod3$$ but 2 is not a square mod 3, so $$z=3r,\quad 2x-y=3s,\quad 6r^2=3s^2-y^2$$ for some integers $r,s$. Then $y$ is a multiple of 3, so $x$ is a multiple of 3, so there is no nonzero solution (because, given any solution, this shows you can divide each variable by 3 to get another solution).
Gerry Myerson
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