# 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 $m\mid n$.

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.

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?

==============================================

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...

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 $$(n,m,\lambda)=(6,3,3),(6,2,4)$$

"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)