In this post that I've asked three weeks ago with same title in Mathematics Stack Exchange and identificator 3692235, for integers $k\geq 1$, we denote the Gregory coefficients as $G_k$. Wikipedia has an article for 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.


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

  • $\begingroup$ Other reference in which I am inspired is On the Diophantine Equation $1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}=g(y)$ by Manisha Kulkarni and B. Sury from the book Diophantine Equations edited by N. Saradha, Tata Institute of Fundamental Research, Narosa Publishing House (2008). $\endgroup$
    – user142929
    Jun 22, 2020 at 18:52
  • $\begingroup$ I've asked my questions, this post, previously in Mathematics Stack Exchange without an answer, but if isn't suitable/interesting for this MathOverflow, please add a commment in next few hours, many thanks. $\endgroup$
    – user142929
    Jun 22, 2020 at 18:52
  • $\begingroup$ On this MathOverflow I've edited the post with title Partial sums involving Gregory coefficients that cannot be an integer. I add this as invitation, that in case that these posts have a good/interesting mathematical content for you, you can visit it and provide a contribution as a comment or adding an answer. $\endgroup$
    – user142929
    Jun 22, 2020 at 19:00
  • $\begingroup$ I've deleted the MSE post. $\endgroup$
    – user142929
    Jun 23, 2020 at 17:20

1 Answer 1


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.

  • $\begingroup$ Many thanks for your excellent answer. I think that it is reasonable that if in next week (the next five or six days) there are no more feedback I should accept your answer. Tomorrow I'm going to study the details of your answer and code. $\endgroup$
    – user142929
    Jun 22, 2020 at 22:50

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