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With Falting's theorem, it can be shown that $$y^2=f(x)$$ where $f(x)$ is a squarefree polynomial with integer coefficients and degree at least $5$ has only finite many rational solutions. Hence, an equation like

$$x^5+x+5=y^2$$ can have only finite many integer solutions. I did neither find a number $p$ such that the equation is not solvable modulo $p$ nor did I find a solution yet.

Can I decide whether such an equation is solvable over the integers in general ? In other words, are efficient bounds known (they need not be unconditional) ?

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  • $\begingroup$ Necessary conditions are : $$x\equiv 4\mod 8$$ $$y\ odd$$ $\endgroup$ – Peter Mar 20 '18 at 18:10
  • $\begingroup$ I think also that any solution $(x, y)$ would also satisfy $x^5 + x + (5 - y^2) = 0$, and so $x \mid 5 - y^2$ by the rational roots theorem. $\endgroup$ – Edward Evans Mar 20 '18 at 18:18
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    $\begingroup$ @ÍgjøgnumMeg $x|y^2-5$ clearly must hold because of $x(x^4+1)=y^2-5$ $\endgroup$ – Peter Mar 20 '18 at 18:21
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Your curve is a hyperelliptic curve of genus $2$. According to the LMFDB:

For a curve $X$ of genus $g\ge 2$ over $\mathbb{Q}$ (or any number fields) the set of rational points $\mathbb{Q}$ is finite, by a theorem of Faltings. At present no algorithm is known that explicitly computes a provably complete list of the points in $X(\mathbb{Q})$, for any given curve $X$, but in specific cases, including when the rank of the Jacobian of $X$ is less then $g$, this can often be accomplished.

The genus of your curve $C$ is greater than the rank of its Jacobian variety $J$. Using the Magma calculator tells us that the genus of $C$ is $2$ and the rank of $J$ is $1$:

> P<x> := PolynomialRing(RationalField());
> C := HyperellipticCurve(x^5+x+5);
> Genus(C);
2
> J := Jacobian(C);
> RankBounds(J);
1 1

Consequently, we can enumerate all the rational points of $C$ using Chaubauty's method. First, note that $C$ has a rational point $I$ at infinity:

> ptsC := Points(C : Bound := 1000);
> I := ptsC[1]; I;
(1 : 0 : 0)

Then check that $J$ has no torsion, i.e., that it is isomorphic to $\mathbb{Z}$:

> TorsionSubgroup(J);
Abelian Group of order 1
Mapping from: Abelian Group of order 1 to JacHyp: J given by a rule [no inverse]

Let's find some points $O$ and $P$ on $J$:

> ptsJ := Points(J : Bound := 1000); ptsJ;
{@ (1, 0, 0), (x^2 + x + 1, 2, 2), (x^2 + x + 1, -2, 2) @}
> O := ptsJ[1]; O;
(1, 0, 0)
> P := ptsJ[2]; P;
(x^2 + x + 1, 2, 2)

Then we check that $O$ is the identity and $P$ has no torsion:

> Order(O);
1
> Order(P);
0

Next, we check that $P$ generates $J$:

> heightconst := HeightConstant(J : Effort:=2, Factor);
> LogarithmicBound := Height(P) + heightconst;  // Bound on h(Q)
> AbsoluteBound := Ceiling(Exp(LogarithmicBound));
> PtsUpToAbsBound := Points(J : Bound:=AbsoluteBound );
> ReducedBasis([ pt : pt in PtsUpToAbsBound ]);
[ (x^2 + x + 1, 2, 2) ]
[1.91676497996095562630531698693]
> Height(P)
1.91676497996095562630531698693

Finally, we invoke Chaubauty's method, thereby proving that $I$ is the only rational point on $C$:

> all_ptsC := Chabauty(P : ptC:=I); all_ptsC;
{ (1 : 0 : 0) }

Consequently, the equation $y^2=x^5+x+5$ has no rational solutions.

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