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I am interested in the congruent number problem which involves finding a rational solution to

$y^2=x^3-x n^2$

This equation is currently unsolved and many have tried. This equation gives the X and the Y values.

However I have a second equation which I believe also must be rational for a solution to exist which is in a different form and not cubic at all. My equation does not give X and Y but rather it gives the D which needs to be multiplied to X and Y to produce an integer triplet where X,Y,Z are all integers.

Once D is known I have a different method to efficiently solve for X,Y,Z.

Where would I look to find the tools to solve 4th order polynomial equations for rational solutions?

My equation is of the form:

$$(c^2)( v^4) - (6 c)( v^2) +1 =k^2$$ Where:
$c$ is a constant given to us.
$V$ and $k$ need to be rational.

Is this easier or harder than a 3rd order equation? Where should I look for the tools to solve this?

Thanks for any help.

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The equation $y^2=x^3-n^2x$ is an elliptic curve, and there is tons known about elliptic curves than most higher dimensional equations. – Thomas Andrews Jun 27 '13 at 2:32
See here:… – Next Jun 27 '13 at 5:57
I was thinking perhaps something like (c v^2-3)^2==L^2+2^2 +2^2 which is in the form " a^2+b^2+c^2 = d^2 {at, bt, ct, dt} = {2(p^2-q^2+r^2), 2(p-q)^2-2r^2-2p(q-r), p^2-(q-r)^2-4r(p-q), 3(p^2+q^2+r^2)-2q(2p+r)} for arbitrary p,q,r and where t is just a scaling factor. ...Proof (Piezas): For any soln a,b,c,d, one can always find rational p,q,r,t using the formulas, " Source But I still am not clear on how to generate rational {V,L} pairs. Seems like this should be something that can be generalized for all c values. – Darrin Taylor Jun 27 '13 at 8:47
An equation of the form $y^2$ equals a quartic is (in general) an elliptic curve and can be massaged into the form $y^2$ equals a cubic, so, in terms of difficulty, the cubic and quartic are equivalent. There are expositions of elliptic curve which tell you how to do the massaging. – Gerry Myerson Jun 27 '13 at 9:27

Note: This is only a partial solution to the $second$ equation.

$(c^2)( v^4) - (6 c)( v^2) +1 =k^2$ is a quadratic in $v^2$. Letting $u = v^2$, this becomes $c^2 u^2 + 6c u+ 1 - k^2=0$.

For this to have a rational root, $36c^2 - 4(c^2 (1-k^2)) = 4a^2(9 + 4k^2-4) =4a^2(5+4k^2)$ must be a square of a rational so $5+4k^2$ must be a square of a rational.

Multiplying out, we want solutions to $5 p^2 + 4q^2 = r^2$.

I'll leave it at this, since I have to go.

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