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I usually work in the field of differential geometry, but I have encountered the following problem in my research: Are there infinitely many positive integers $k,l,m\in\mathbb N^{>0}$ such that $$(3+3k+l)^2=m\,(k\,l-k^3-1)\,?$$ Obviously, taking $l=k^2$ and $m=-(3+3k+l)^2$ gives infinitely many integer solutions, but $m<0$ is negative. As a non-expert, I imagine that there is either a simple answer to this question, or the problem is not so simple to solve. Of course, I've played around with the equations a bit, but other than finding numerous examples, I haven't made any progress.

I would appreciate an existence or non-existence statement for infinitely many positive integer solutions, but also some hints that the problem is most likely hard to solve would help me.

Background: I am looking for certain integer representations of a surface group, and I can show that integer solutions to this diophantine equation actually give rise to integer representations. The condition that $k,l,m$ are positive is equivalent to the condition that the corresponding representation is contained in a higher Teichmüller component (which is important for my differential geometric application).

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    $\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. $\endgroup$
    – GH from MO
    Jan 21, 2022 at 15:32
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    $\begingroup$ Writing $k = l^2+d$ reduces the problem to $(k^2+3k+3+d)^2 \equiv 0 \pmod{kd-1}$, and then multiplying each side by $k^2$ (which is relatively prime to $kd-1$, so no extra solutions due to this), noting $kd \equiv 1 \pmod{kd-1}$, and using $(k+1)^3 = k^3+3k^2+3k+1$, one sees that your problem is equivalent to finding infinitely many positive integers $k,d$ for which $$kd-1 \mid (k+1)^6.$$ $\endgroup$ Jan 21, 2022 at 16:44
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    $\begingroup$ @mathworker21: From this perspective, the series from my answer gives divisibility $(kd-1)\mid (k+1)^4$. I wonder if for some other series the degree of $k+1$ can be lowered even further. $\endgroup$ Jan 21, 2022 at 22:14
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    $\begingroup$ @MaxAlekseyev My code is not finding any $(k,d) \in \mathbb{N}$ with $k \ge 15$ for which $(kd-1) \mid (k+1)^3$. $\endgroup$ Jan 21, 2022 at 22:25
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    $\begingroup$ @MaxAlekseyev Here is a proof that there are finitely many pairs $(k, d) \in \mathbb{N}$ such that $k d - 1 | (k + 1)^3$: multiplying the right hand side by $d^3$ and replacing $k d$ with $1$, this is equivalent to $k d - 1 | (d + 1)^3$, therefore WLOG $k \leq d$. Now, if $(k + 1)^3 = m (k d - 1)$, then since $m = -1 (\bmod k)$ we have $m \geq k - 1$ and in fact $m = k - 1$ can happen only finitely often, therefore we can assume $m \geq 2 k - 1$. Thus, $(k + 1)^3 = m (k d - 1) \geq (2 k - 1) (k^2 - 1)$, which can happen for only finitely many $k$. $\endgroup$
    – Random
    Jan 21, 2022 at 22:55

2 Answers 2

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It does have infinitely many positive solutions. Here is just one such series.

Consider the following recurrence sequence:

$$u_0=1,\ u_1=2,\ u_{n+1} = 23 u_n - u_{n-1} - 4\qquad (n\geq 1).$$

Let $t,k$ be any two consecutive terms of this sequence, then setting $l:=k^2+t$ produces the following equality: $$(3+3k+l)(t+1) = (k+26)(kl-k^3-1),$$ which gives solution $m:=\frac{(k+26)(3+3k+l)}{t+1}$ (which is an integer) to the original equation.


In fact, integrality of $m$ follows from the identity: $$(u_{n+2}+1)(u_n+1) = (u_{n+1}+26)(u_{n+1}+1),$$ which can be verified from the recurrence for $u_n$.

In summary, the values $(k,l,m)$ in this solution series are given by $$\begin{cases} k = u_{n+1}, \\ l = u_{n+1}^2 + u_n, \\ m = (u_{n+2}+2)(u_{n+1}+2) + 24. \end{cases}\qquad (n\in\mathbb{Z}_+) $$


ADDED. I've added $u_n$ to the OEIS as sequence A350917. Together with 9 other similar recurrences it gives all solutions $k$ to $(tk-1)\mid (k+1)^4$, which are now listed in sequence A350916.

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    $\begingroup$ Hi Max, your solution is great, many thanks. May we use your proof in our article. Of course, we will give you full credit for it. $\endgroup$
    – Sebastian
    Jan 22, 2022 at 7:54
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    $\begingroup$ @Sebastian: Sure. Btw, I've classified all solutions to $(tk-1)\mid (k+1)^4$ (which are given by 10 recurrences including the one from my answer) and believe it can also be done for $(tk-1)\mid (k+1)^6$ (i.e. obtaining all solutions to your equation). Let me know if you are interested. $\endgroup$ Jan 22, 2022 at 13:56
  • $\begingroup$ What is the proof for your classification of all the solutions to $(tk-1)\mid(k+1)^4$? $\endgroup$
    – user178594
    Mar 25, 2023 at 14:49
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    $\begingroup$ @JovanRadenkovic: $(tk-1)\mid (k+1)^4$ is equivalent to $(tk-1)\mid (k^2 + t^2 + 4k + 4t + 6)$, which can be solved with Vieta jumping. $\endgroup$ Mar 25, 2023 at 15:37
  • $\begingroup$ @MaxAlekseyev, can this method be used for an exponent greater than $4$? $\endgroup$
    – user178594
    Mar 25, 2023 at 16:27
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Here is another way maybe simpler as it reveals a certain Pell-Fermat equation. I don't know the relation with the other answer but i did many restrictions to get the following infinite sequence of solutions:

For $n\ge 1$, $$\begin{cases}q_n=-7q_{n-1}+24d_{n-1}\\d_n=2q_{n-1}-7d_{n-1}\end{cases}$$ with $q_0=3$ and $d_0=1$, (solutions of $q^2-12d^2=-3).$ Set $j=2|d_n|$, $i=4|d_n|+|q_n|$, $r=3(i+j)$, so that $m=r^2$, $k=ij-1$ and $x:=3+3k+l=\dfrac{mk+r|j^3-i^3|}{2}$. This implies the positive integer solutions $(m,k,l)$ are infinite. The starting equation to solve in $ x $ over $\mathbb{N}$ was $x^2-mkx+m(k+1)^3$.

Its discriminant is a square $X^2=(r\cdot y)^2$ as $\Delta=X^2=m^2k^2-4m(k+1)^3$, say $m=r^2$ is a perfect square, then $(rk-y)(rk+y)=4(k+1)^3$ and set $(k+1)^3=i^3j^3$, ($k=ij-1$). $$\begin{cases}rk-y=2i^3\\rk+y=2j^3\end{cases}.$$ So $r=(i+j)(-1+\dfrac{i^2+j^2-1}{ij-1})$ and $y=j^3-i^3$. Let $\dfrac{i^2+j^2-1}{ij-1}=s$ an integer say $s=4$. Finally we solve in $i$, $i^2+j^2-1=4ij-4$, taking again its discriminant to be a perfect (even) square $p=2q$, $j=2d$ we solve $p^2-12j^2=-12$ as $q^2-12d^2=-3$.

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    $\begingroup$ This solution series is unrelated to my answer since the power of $k+1$ in divisibility here is 6. Btw, both $j$ and $i$ are recurrence sequences satisfying $a_n = 14a_{n-1} - a_{n-2}$ for $n\geq 3$. They are present in the OEIS as A094347 and A011943, respectively. Also, they may be viewed as every other pair of consecutive terms of A001075. The other pairs from this sequence, like $(7, 26)$ with $k=181$, also give solutions. $\endgroup$ Jan 23, 2022 at 15:16
  • $\begingroup$ In fact, for the terms of A001075 we have $a_n^6 \equiv a_{n+1}^6 \equiv -1 \pmod{a_na_{n+1}-1}$, proving that $k:=a_na_{n+1}-1$ is a solution to $(kt-1)\mid(k+1)^6$. These values of $k$ are given in A001570. $\endgroup$ Jan 23, 2022 at 15:51

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