# Does this equation have any nonzero solutions

Does the equation

$$(a^{2}-b^{2})^{4}(a^{2} + b^{2}) = 4a^{2}b^{2}$$

Have any nonzero solutions, where $a^2$ and $b^2$ are both rational ?

If yes, i suspect there should be infinitely many of them, can they be parametrized ?

• Just to clarify: $a^2$ and $b^2$ need to be rational, but $a$ and $b$ need not be rational? In that case, why don't you write your equation as $(x-y)^4(x+y)=4xy$? – RP_ Oct 11 '16 at 11:34
• Why are you interested in this equation and its generalization? – Michael Stoll Oct 11 '16 at 13:48

In any case there are only finitely many rational solutions. If I interpret your question correctly, you are looking for rational points on the curve $C$ given by $$(x-y)^4(x+y)=4xy$$ where I have put $x=a^2$ and $y=b^2$. Taking $t = x/y$, we easily find that $y$ satisfies $$y^3 = \frac{4t}{(t-1)^4(t+1)},$$ so that $C$ is birational to the superelliptic curve $C'$ given by $$\eta^3 = 4 \xi (\xi-1)^2 (\xi + 1)^2.$$ We thus see that $C'$ is a triple cover of $\mathbb{P}^1$ which is totally ramified in $4$ points (including one above infinity), so by Riemann--Hurwitz $C'$ has genus $2$, so has finitely many rational points.

• Just curious: how do you get the third display from the second? That is, what is the birational map $(x,t)\to(\eta,\xi)$? – GH from MO Oct 11 '16 at 12:28
• I think it is simply $\eta = y(t-1)^2(t+1)$ and $\xi = t$. – RP_ Oct 11 '16 at 12:30
• Thanks. Also, in the second display, $x^3$ should really be $y^3$. – GH from MO Oct 11 '16 at 12:32
• @Rene, thank you very much for your answer. Would you also please consider the more general equation: $$(x-y)^{4n}(x+y) = 4^{n}x^{2n}y^{2n}$$ where $x,y$ are rational and $n\geq 1$ is an integer ? – PNT. Oct 11 '16 at 13:15
• Do you mean $(x-y)^{4n}(x+y)=4^n x^n y^n$? Otherwise it does not have your original problem as a special case. – RP_ Oct 11 '16 at 13:25

Extending René's answer, the curve $C'$ is birational to $$C'' \colon y^2 = x^6 + 4,$$ whose Jacobian variety has finite Mordell-Weil group isomorphic to $({\mathbb Z}/3{\mathbb Z})^2$, from which it is easy to show that the only rational points on $C''$ are thoses with $x = 0$ and the two points at infinity. From this, it is easy to get all solutions to the original equation.

To get to this hyperelliptic equation, we rewrite René's equation as $$(\xi^2 - 1) \left(\frac{\eta}{\xi^2-1}\right)^3 - 4 \xi = 0$$ and set $u = \eta/(\xi^2-1)$. Taking the discriminant of $u^3\xi^2 - 4\xi - u^3$ as a polynomial in $\xi$ gives $$4 + u^6 = v^2$$ as claimed. The points with $u = 0$ give $\eta = 0$, so $y = 0, x, -x$, which leads to $x = y = 0$. The points at infinity have $\xi = \pm 1$, which gives the same result.

• thank you very much for your answer. Would you also please consider the more general equation: $$(x-y)^{4n}(x+y) = 4^{n}x^{2n}y^{2n}$$ where $x,y$ are rational and $n\geq 1$ is an integer ? – PNT. Oct 11 '16 at 13:14
• Typo in my above comment: the factor on the RHS should be $x^{n}y^{n}$, not $x^{2n}y^{2n}$. – PNT. Oct 11 '16 at 13:28
• For larger $n$, the curves will no longer be hyperelliptic, so the machinery that I used to deal with $n = 1$ is no longer applicable. – Michael Stoll Oct 11 '16 at 13:49