Consider the equation $a^4+6v^2a^2-8a+v^4=0$ over the rationals. Note that the following are solutions: $(a,v)=(1,1),(0,0),(2,0)$. Are there any other rational solutions?
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2$\begingroup$ Not sure if it helps, but this can be rewritten as $(a^{2} + 3v^{2})^{2} = 8(v^{4}+a).$ $\endgroup$– Geoff RobinsonCommented Jan 27, 2020 at 11:01
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$\begingroup$ Thank you very much for your comments.My general question is how we can determine the coefficients of a quartic equation(or higher degree eqtation) ax^4+bx^+cx^3+dx^2+e=0 such that it has rational solutions. A trivial method is first determine the roots (x_i), then the coefficients!!!... I mean as example how I determine the coefficients a,b in the quartic equation 3x^4+2^x^3+ax^2+x+ b=0 such that it has the rational roots? $\endgroup$– user151553Commented Jan 27, 2020 at 11:28
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1$\begingroup$ @user151553: There is no general algorithm to decide if a given quartic equation has a rational solution. This follows from Matiyasevich's theorem, because every system of polynomial equations can be encoded into a single quartic equation (using sums of squares of quadratic equations, including simple variable changes like $x=yz$). $\endgroup$– GH from MOCommented Jan 27, 2020 at 13:26
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$\begingroup$ For $y=v^2+3a^2$, the curve $y^2= 8a^4+8a$ is an elliptic curve; so it should not be hard to find its rank. If $0$, the answer is easy, otherwise one needs harder method for genus 3 curves. What is the motivation for the question? $\endgroup$– Chris WuthrichCommented Jan 27, 2020 at 15:24
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$\begingroup$ @ChrisWuthrich: That curve has rank 1. See my comment below Luis Ferroni's answer. $\endgroup$– GH from MOCommented Jan 27, 2020 at 15:32
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1 Answer
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This is not a complete solution but rather a line of action. I think this method may work completely, but it is too tedious to make step by step.
Note that you can think of your equation as a quadratic with variable $v^2$. Said that, for $v^2$ to be rational, the discriminant of the quadratic has to be the square of a rational number, this is:
$$36a^4 - 4(a^4-8a)=\left(\frac{m}{n}\right)^2$$ $$2a(a^3+1) = \left(\frac{m}{4n}\right)^2$$
Writing $a=\frac{p}{q}$ with $\gcd(p,q)=1$, and replacing above:
$$2p(p^3+q^3) = \left(\frac{q^2m}{4n}\right)^2$$
Which in turn, since the LHS is an integer number, implies that $4n$ divides $q^2m$. Notice that:
$$d:=\gcd(p,p^3+q^3) = \gcd(p,q^3) = 1$$
In particular $d:=\gcd(2p,p^3+q^3)\in \{1,2\}$.
If $d=1$, then one has that $2p$ and $p^3+q^3$ have both to be squares of integers. Furthermore, notice that $p^3+q^3=(p+q)(p^2-pq+q^2)$, and also
$$d':=\gcd(p+q,p^2-pq+q^2) \in \{1,3\}$$
In any case, one has that $\frac{p+q}{d'}$ and $\frac{p^2-pq+q^2}{d'}$ have to be squares of integers. A bit of manipulation converts the equation $p^2-pq+q^2=u^2d'$, into the Pell equation with variables $x,y\in\mathbb{Q}$: $$ x^2 + 3y^2 = d'$$ where $x=\frac{2p-q}{2u}$ and $y=\frac{q}{2u}$. If $d'=1$ it can be solved by standard methods. If $d'=3$ it is very similar.
Once you've got the solutions of these Pell equations, you can parametrize all possible values of $p$ and $q$. Plugging them into our conditions that $\frac{p+q}{d'}$ and $2p$ had to be squares of integers, we will probably get a complete parametrization of the feasible couples $(p,q)$. This gives you all possible values of $a$ for this case.
From them you get all possible rational values of $v^2$.
Its similar to Case 1, but with $d=2$. The steps should be similar (probably this one will require more care).
answered Jan 27, 2020 at 14:48
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1$\begingroup$ SAGE tells that your second display defines an elliptic curve, which is isomorphic to the one defined by $y^2=x^3+8$. SAGE also tells that this elliptic curve has infinitely many rational points. Relevant SAGE commands: "a, b = polygen(QQ, 'a, b')", "E = Jacobian(b^2-2*a^4-2*a)", "print E", "print E.mwrank()". $\endgroup$ Commented Jan 27, 2020 at 15:12
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$\begingroup$ Thank you for the above comments. But I do not understsnd it completely..Does it yield a solution to the problem? $\endgroup$ Commented Jan 28, 2020 at 11:27
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1$\begingroup$ Well I only guided you through a path that can be useful (at least to try to find other non-trivial solutions). It's up to you to fill the details and see if it works. $\endgroup$ Commented Jan 28, 2020 at 13:04