Find solutions of $m=\frac{n^2}{(n-m-1)\lambda+n}$ where $n,m,\lambda$ are postive integers,$1\le\lambda \le n-1$ and $m\mid n$.

Question

I am considering the following equation $$m=\frac{n^2}{(n-m-1)\lambda+n}$$ where $n,m,\lambda$ are postive integers, $1\le\lambda \le n-1$ and $m\mid n$. If $m=n$, then $$\frac{n^2}{(n-n-1)\lambda+n}=\frac{n^2}{n-\lambda}\ne n.$$ So we can assume that $m<n$.

clear;
for m in [1..100] do
for k in [2..100] do
  n:=k*m;
for r in [1..(n-1)] do
  p:=n^2;
  q:=(n-m-1)*r+n;
  if p mod q eq 0 then
  if p div q eq m then
    print r,m,n,p,q;
  end if;
  end if;
end for;
end for;
end for;

I run the above Magma code and it outputs

4 2 6 36 18
3 3 6 36 12

So I conjecture that

Except the above two example, i.e., $(n,m,\lambda)=(6,2,4)$ and $(n,m,\lambda)=(6,3,3)$, there does not exist three positive integers $n,m,\lambda$ such that $1\le\lambda \le n-1$, $m\mid n$ and $$m=\frac{n^2}{(n-m-1)\lambda+n}.$$

Can anyone proof the conjecture or give a counterexample?

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My trying: For $\lambda=1$, I can proof that there exist two positive integers $n,m$ such that $m\mid n$ and $m=\frac{n^2}{2n-m-1}$.

Proof. Assume that there exist two positive integers $n,m$ such that $m\mid n$ and $m=\frac{n^2}{2n-m-1}$. Then we have $$ m(2n-m-1)=n^2\tag{1}$$ and $$n=km\tag{2}$$ where $k$ is an integer. Put $\mathrm (1)$ into $\mathrm (2)$ and obtain $$m(2km-m-1)=k^2m^2\tag{3}.$$ Since $m$ is positive, dividing both sides of $\mathrm (3)$ by $m$ and we have $$(2k-1)m-1=k^2m\tag{4}.$$ Obviously, the right hand side of $\mathrm (4)$ is a multiple of $m$ while the other side is not. It is a contradiction. So there could not exist such two integers $n$ and $m$. QED.

In a similar way, we can proof that if such integers exist, then $m\mid \lambda$. To be continued...

Answer 1

The conjecture is true.

Proof :

Let $n=mk$ where $k\ge 2\in\mathbb Z$. Then, we have $$(mk-m-1)\lambda=mk(k-1)$$ If $m=1$, then $\lambda=\frac{k(k-1)}{k-2}=k+1+\frac{2}{k-2}$. So, we have to have $k-2\mid 2$ implying $k=3,4$. So, we have $(n,m,\lambda)=(3,1,6),(4,1,6)$, but these don't satisfy $\lambda\le n-1$.

If $m\ge 2$, then since $m\not\mid mk-m-1$, we can write $\lambda=ma$ where $a$ is a positive integer such that $a\le k-\frac 1m$, i.e. $a\le k-1$. Then, we have

$$ma(k-1)-a=(k-1)k\implies a=(k-1)(ma-k)$$ from which we have $ma-k=1$ and $a=k-1$. So, eliminating $k$, we have $a(m-1)=2$.

• $(a,m-1)=(1,2)$ implies $(n,m,\lambda)=(6,3,3)$

• $(a,m-1)=(2,1)$ implies $(n,m,\lambda)=(6,2,4)$

Therefore, the only solutions are $$(n,m,\lambda)=(6,3,3),(6,2,4)$$

Answer 2

"OP" requested a counter-example numerical solution:

(n,m,y)=(6,3,5)

above satisfies the given equation

& also y=(n-1)=5 & (m divides n)