Find all natural solutions $m!=a^2n!$ .  
   

It's clear that $m>=n$.  
 
When $m=n$ we have solutions $(1,m,m)$.  
 
When $m=n+1$ we have solutions $(a,a^2,a^2-1)$.  
 
I think that when $m>n+1$ we have no solutions but I can't prove this.



Diophantine equation with factorials
[The product of consecutive integers is never a power.](https://projecteuclid.org/euclid.ijm/1256050816)


