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I have been working on solutions to $x^5+y^5+z^5=1$, and I found that the three solutions of $x^3+bx+\frac{1}{5b}$ satisfy that equation. Multiplying by $5b$: $5xb^2+5x^3b+1=0$, then solving for b yields:

$b=\frac{-5x^3+\sqrt{25x^6-20x}}{10x}$

Which leads me to my question... What are the rational solutions of the diophantine equation: $y^2=25x^6-20x$?

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    $\begingroup$ Just a small remark that you are probably aware of: there are only finitely many rational solutions to your diophantine equation as it is a hyperelliptic curve of genus 2 (see Faltings' theorem). $\endgroup$ – GH from MO May 17 '18 at 0:47
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    $\begingroup$ mathoverflow.net/questions/226093/… $\endgroup$ – individ May 17 '18 at 4:35
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    $\begingroup$ Probably just the point with $x=0$ and the two points at infinity. But the Jacobian has rank 2 so it's not yielding to off-the-shelf methods. $\endgroup$ – Felipe Voloch May 17 '18 at 7:11
  • $\begingroup$ I knew it would only have finitely many solutions, I was hoping that there would be a nontrivial case which could lead to a solution or at least a very interesting special case. The above cubic corresponds to the solutions where x+y+z=0. The case x+y+z=1 corresponds to the parametric curves in individ's and my answer in the linked question, which are both unfortunately complex (although my solution allows one of the values to be real). Any non-off-the-shelf methods I could try? $\endgroup$ – Thomas May 17 '18 at 12:20
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I probably put a little bit too much effort into this. The only rational point on this curve is $(0,0)$ (as well as the points at infinity $(1 : 5 : 0)$ and $(1 : -5 : 0)$).

There's a slightly non-obvious change of variables that will turn your equation into $y^{2} = x^{5} + 2000^{2}$. Using this form of the equation, we can lift the points on the curve to one of a family of étale $5$-fold covers. In particular, if we write $x = a/b$ and $y = c/d$ with $\gcd(a,b) = \gcd(c,d) = 1$, then we get $$ \frac{c^{2}}{d^{2}} = \frac{a^{5} + 2000^{2} b^{5}}{b^{5}}. $$ It's easy to see that the fraction on the right is reduced and so $c^{2} = a^{5} + 2000^{2} b^{5}$ and $d^{2} = b^{5}$. Thus, $b = n^{2}$ for some positive integer $n$. This gives $$c^{2} - 2000^{2} n^{10} = (c - 2000 n^{5})(c + 2000 n^{5}) = a^{5}.$$ The two factors on the left hand side can only have the prime factors $2$ and $5$ in common, and their product is a $5$th power. Hence, $c + 2000n^{5} = 2^{\alpha} 5^{\beta} e^{5}$ for some integers $0 \leq \alpha, \beta \leq 4$. Letting $f = a/e$, we get $$ 2^{2 \alpha} 5^{2 \beta} e^{5} = f^{5} + 2^{\alpha + 5} 5^{\beta + 3} n^{5}. $$ Each of these $25$ curves has genus $6$, and every rational point on the original curve lifts to one of these $25$ curves. The only ones that have local points are the three with $(\alpha,\beta) = (0,0), (0,2)$ and $(0,3)$. The curve with $(\alpha,\beta) = (0,0)$ is $a^{5} + b^{5} = 125c^{5}$, and the curves with $(\alpha,\beta) = (0,2)$ and $(0,3)$ are both isomorphic to $a^{5} + b^{5} = 625c^{5}$. The three rational points on the original curve all arise from the point $(1 : -1 : 0)$ on these curves. It suffices to prove there are no more points on these curves, in particular that it's not possible to write $125$ and $625$ as a sum of two rational $5$th powers.

I'll do this by using Frey curves. (There's a more elementary way to do the case of $a^{5} + b^{5} = 125c^{5}$, but the genus $2$ quotient of $a^{5} + b^{5} = 625c^{5}$ also has a Jacobian with $\mathbb{Q}$-rank $2$.)

Suppose that $a$, $b$ and $c$ are integers with $c \ne 0$ and $\gcd(a,b,c) = 1$ so that $a^{5} + b^{5} = 5^{\beta} c^{5}$ with $\beta \in \{3, 4 \}$. It's easy to see that $5 \nmid a$ and $5 \nmid b$. First assume that $c$ is odd. Swap $a$ and $b$ if necessary to force $a$ to be odd, and negate $a$ if necessary to make $a^{5} \equiv -1 \pmod{4}$. Then, $b$ is even. Let $E$ be the elliptic curve $$E : y^{2} = x(x-a^{5})(x+b^{5}).$$ The discriminant of this curve is $16 \cdot 5^{2 \beta} a^{10} b^{10} c^{10}$, although there is a change of variables that will lower the discriminant and show that $E$ has multiplicative reduction at $2$. The $j$-invariant of $E$ is $\frac{(5^{2 \beta} c^{10} - a^{5} b^{5})^{3}}{5^{2 \beta} a^{10} b^{10} c^{10}}$.

Thinking about the mod $5$ Galois representation attached to $E$ will lead to a contradiction. This representation is irreducible. (All the $2$-torsion points on $E$ are rational, and if the mod $5$ representation was reducible, this would give a rational point on the modular curve parametrizing elliptic curves with full $2$-torsion and reducible mod $5$ Galois representation. This modular curve has genus $1$, and precisely $6$ rational points, all of which are cusps.)

Serre's conjecture (which is now a theorem) predicts that there is a modular form of weight $k$ and level $N$ whose mod $5$ Galois representation gives rise to that of $E$. The level $N$ is the product of the primes $\ell \ne 5$ for which the power $\ell$ dividing the discriminant is not a multiple of $5$. By the computation above, this is only $\ell = 2$, so $N = 2$. The weight of the modular form is $6 = 5+1$, because the power of $5$ dividing the discriminant is not a multiple of $5$. However, we get a contradiction because there are no weight $6$ cusp form of level $2$! (The lowest weight cusp form of level $2$ arises in weight $8$.) This contradiction proves that this elliptic curve doesn't exist, and so there are no solutions to $a^{5} + b^{5} = 5^{\beta} c^{5}$ with $c$ odd.

In the case that $c$ is even and $a$ and $b$ are odd, one can run the same argument starting with the elliptic curve $E : y^{2} = x(x-a^{5})(x-5^{\beta}c^{5})$ and the same things happen - the irreducible mod $5$ Galois representation arises in level $2$ and weight $6$, which is again a contradiction.

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