Can you solve this?

Suppose $\frac{n^3+1}{mn-1}=k$ where $k$ is a positive integer. Then $n^3+1=(mn-1)k$ and so it is clear that $k\equiv -1\pmod{n}$ . So, let $k=jn-1$ where $j$ is a positive integer. Then we have $n^3+1=(mn-1)(jn-1)=mjn^2-(m+j)n+1$ which by cancelling out the 1s and dividing by $n$ yields $n^2=mjn-(m+j)\Rightarrow n^2-mjn+m+j=0$ . The equation $x^2-mjx+m+j=0$ is a quadratic. We are given that $n$ is one of the roots. Let $p$ be the other root. Notice that since $n+p=mj$ we have that $p$ is an integer, and so from $np=m+j$ we have that $p$ is positive.

It is obvious that $j=m=n=p=2$ is a solution. Now, if not, and $j,m,n,p$ are all greater than $1$ , we have the inequalities $np>n+p$ and $mj>m+j$ which contradicts the equations $np=m+j, n+p=mj$ . Thus, at least one of $j,m,n,p$ is equal to $1$ .

If one of $m,j$ is $1$ , without loss of generality assume it is $j$ . Then we have $np=m+1, n+p=m$ . That is, $np-n-p=1\Rightarrow (n-1)(p-1)=2$ which gives positive solutions $(n,p)=(3,2),(2,3)$ . These give $m=5$ and since we assumed $j=1$ , we can also have $m=1$ and $j=5$ .

If one of $n,p$ is $1$ , without loss of generality assume it is $p$ . Then we have $n=m+j, n+1=mj$ . That is, $mj-m-j=1\Rightarrow (m-1)(j-1)=2$ which gives positive solutions $(m,j)=(3,2),(2,3)$ . These give $n=5$ and since we assumed $p=1$ , we can also have $n=1$ and $p=5$ .

From these, we have all solutions $(m,n)=(2,2),(5,3),(5,2),(1,3),(1,2),(3,5),(2,5),(3,1),(2,1)$ . Hence we have a total of 9 ordered pairs.

Hey, I like this problem.

First note that if $mn-1|n^3+1$ then $mn-1|m^3n^3-1+n^3+1=n^3(m^3+1)$ It follows that $mn-1|m^3+1$ because $\gcd(mn-1, n^3)=1$ , so if $(m,n)$ is an answer $(n,m)$ will be too. If $n=m$ we get $\frac{n^3+1}{mn-1}=\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1}$ So $n-1|1$ and $n=2$ . Therefore $(2,2)$ is an answer and from now on we can suppose $m>n$ . Let $n=1$ , it gives $m=2,3$ , we get four answers from here: $(1,2),(2,1),(1,3),(3,1)$ Note that if $\frac{n^3+1}{mn-1}=r$ then $r\stackrel{n}{\equiv} -1$ , and $r$ have the form of $kn-1$ for some positive integer $k$ . Since $m>n$ $kn-1=\frac{n^3+1}{mn-1}<\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1} \\ kn-1<n+\frac{1}{n-1} \\ k<1+\frac{1}{n}+\frac{1}{n(n-1)} \le 2 \\ k<2$ Thus $k=1$ and we have $(n-1)(mn-1)=n^3+1$ , hence $n-1|n^3+1$ or $n-1|n^3+1-(n^3-1)=2 \\ n-1|2$ As a result $n=2$ or $n=3$ which gives us $2m-1=9$ or $2(3m-1)=28$ and both of them leads to $m=5$ , so we get four other answers from here: $(2,5),(5,2),(3,5),(5,3)$ .

I'm going to hazard a guess that there are no solutions where $m$ or $n$ is greater than $1000.$ Then this is actually very simply solved by checking all the ordered pairs $(m, n)$ for positive integers $m$ and $n$ under 1000.

public class CubicSquareRatio
{
    public static void main(String[] args)
    {
        for (double length = 5.0; length < 1000.0; length++)
        {
            int count = 0;
            for (double m = 1; m < length; m++)
                for (double n = 1; n < length; n++)
                {
                    if (((n * n * n + 1.0)/(m * n - 1.0)) % 1.0 == 0.0)
                        count++;
                }
            System.out.println(count);
        }
    }
}

Indeed, it prints " $5$ " for length = 5.0, and " $9$ " for every length from 6.0 to 999.0. Unless there is some solution with $m \geq 1000$ or $n \geq 1000$ (in which case there would probably be infinitely many solutions), there are only the $\boxed{9}$ solutions found.

Note that $(m, n)$ is a solution iff

$\large\ { m }^{ 2 }{ n }^{ 2 } + mn + 1 + \frac { { n }^{ 3 } + 1 }{ mn - 1 } = \frac { { n }^{ 3 }({ m }^{ 3 } + 1) }{ mn - 1 }$

is an integer. Since $gcd(n, mn - 1) = 1$ , this occurs iff $\frac { { m }^{ 3 } + 1 }{ mn - 1 }$ is a solution. That is, $(m, n)$ is a solution iff $(n, m)$ is a solution. Suppose $(m, n)$ is a solution and $mn > 1$ ; defining $q$ as in the current solution, we see that $q<n$ . Further, since
${ ({ n }^{ 2 }\quad -\quad 1) }^{ 2 }\quad \ge \quad { n }^{ 3 }\quad +\quad 1$ for $n \ge 2$ , we see that $q < n$ . Similarly, if $m = n > 2$ , we get $q < n$ .

Thus the following steps

$(i)$ If $m < n$ , interchange $m$ and $n$ ,\

$(ii)$ If $m > n > 1$ or $m = n > 2$ , replace $(m, n)$ by $(q, n)$ ,

starting from any solution we can always reduce to a smaller solution untill we get to either $m = n = 2$ or $n = 1$ .

In the latter case we have $m = 2 {or} 3$ . Backtracking gives the chain of solutions

$(3,5(\rightarrow \quad (5,3)\quad \rightarrow \quad (1,3)\quad \rightarrow \quad (3,1),\quad (2,5)\quad \rightarrow \quad (5,2)\quad \rightarrow \quad (1,2)\quad \rightarrow \quad (2,1)$ and the single solution $(2, 2)$ .

Thus these $9$ pairs are all solutions to the problem.

Answer: $\boxed{9}$ .

Python

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N=0
for n in range(2,999):
    for m in range(1,999):
        if int((n**3+1.0)/(m*n-1))==(n**3+1.0)/(m*n-1):
            N+=1
                print N,[m,n]