In this post that I've asked three weeks ago with same title in [Mathematics Stack Exchange](https://math.stackexchange.com/questions/3692235/diophantine-equations-that-involve-gregory-coefficients-a-computational-exercis) and identificator **3692235**, for integers $k\geq 1$, we denote the Gregory coefficients as $G\_k$. Wikipedia has an article for [*Gregory coefficients*](https://en.wikipedia.org/wiki/Gregory_coefficients), are known as reciprocal logarithmic numbers (I add this as additional reference). I was inspired in problems that I know from the literature (in particular [1], that is from the section of problems of a journal) to solve the following diophantine equation that involves the first few Gregory coefficients in the brackets from RHS
$$y^2=1+\left(\frac{1}{2}n-\frac{1}{12}n^2+\frac{1}{24}n^3\right)$$
where we consider that $y\geq 1$ is integer and $n\geq 1$ also is integer.



> 
> **Question 1.** Prove or refute that the previous diophantine equation $$y^2=1+\sum\_{k=1}^3G\_k \cdot n^k\tag{1}$$
> have no solutions $(n,y)$ when $y\geq 1$ and $n\geq 1$ run over positive integers. Can you find a counterexample? **Many thanks.**
> 
> 
> 


My claim here was the following, that summarizes the things that I can see here (I don't know if previous question is easy to get). Also I know that $(1)$ is an elliptic curve (but in this post I'm interested in integral solutions).


**Claim.** *Our equation* $(1)$ *can be rewritten as* $n((n-2)n+12)=24(y-1)(y+1)$ *(with help of Wolfram Alpha online calculator). From here we get easily (by contradiction) than* $n$ *is an even integer. And* $n\equiv 0\text{ mod }3$ *or* $n\equiv 2\text{ mod }3$.


I've tested the conjecture stated in previous question for humble sets of integers. On the other hand I'm curious if there is some diophantine equation of the form $y^2=1+\sum\_{k=1}^ NG\_k n^k$ for some integer $N>3$ for which we can compute at least an integral solution $(n,y)$.



> 
> **Question 2 (A computational exercise).** Can you show an example of diophantine equation $$y^2=1+\sum\_{k=1}^N G\_k \cdot n^k\tag{2}$$ with at least a solution $(n,y)$, for integers $n,y\geq 1$ as before, where $N>3$? **Many thanks.**
> 
> 
> 


I tried with my computer the first few values of $N$, the lowest of these integers $N>3$, and for $1\leq n,y\leq 5000$ both integers. If you can to answer **Question 2** with a family of integral solutions, or you can find several examples of $N$ for diophantine equations $(2)$ having solutions feel free to expand your answer of this question.


I don't know if my questions are in the literature. If you know some of these from the literature refer it ansewring the questions as a reference request.


References:
-----------


[1] Fuxiang Yu, *An Old Fermatian Problem: 11203*, Problems, The American Mathematical Monthly, Vol. 114, No. 9 (Nov., 2007), p. 840.



Diophantine equations that involve Gregory coefficients: a computational exercise
Integral points on elliptic can often be computed routinely. In Question 1, the curve can be rewritten as
$$Y^2 = 5184 + 432 X -12X^2 + X^3,$$
where $X:=6n$ and $Y:=72y$. SageMath computes:



```
sage: EllipticCurve([0,-12,0,432,5184]).integral_points()
[(0 : 72 : 1), (21 : 135 : 1)]

```

So, the only integer solution is $(n,y) = (0,1)$.


For Question 2 with $N=4$, we get a hyperelliptic curve:
$$(60y)^2 = 3600 + 1800n -300n^2 + 150n^3 -95n^4,$$
where the integral points can be found by Magma:



```
> IntegralQuarticPoints([-95, 150, -300, 1800, 3600]);
[
 [ 0, 60 ]
]

```

So, again $(n,y) = (0,1)$ is the only integer solution.


This does not rule out the possibility of non-trivial solutions for $N>4$, but at very least we know that for any fixed $N>4$ there are only finitely many solutions.


