Find all pairs (m, n) of positive integers such 
that m²/(2mn² - n³ + 1) is a positive integer. 

background of the problem ..some preliminary concepts and review of
related literature


       m²
———————————————
 2mn² - n³ + 1

Case 1.  When the denominator = 1, then the above fraction is m², which
is a positive integer

The denominator = 1 is true when and only when

 2mn² - n³ + 1 = 1, which is true when and only when
     2mn² - n³ = 0, which is true when and only when
    n²(2m - n) = 0

n is not 0 since it is a positive integer, so the last
equation is true when and only when

        2m - n = 0, which is true when and only when
            2m = n

So case 1 tells us that the given expression is a positive
integer when n = 2m.  So the ordered pair (2m,m) will always
result in a positive integer for the given expression.

---------------------------------------

Case 2.  When n = 1, the last two terms cancel

       m²
——————————————— =
 2mn² - n³ + 1         

         m²
——————————————————— =
 2m(1)² - (1)³ + 1    
       
        m²
       ———— =
        2m

         m
        ———
         2

And if m is chosen even, say m = 2k,
then the expression will be an integer,

So pairs (m,n) = (2k,1)

----------------------------------------

When n = 2

       m²
——————————————— =
 2mn² - n³ + 1 

         m²
——————————————————— =
 2m(2)² - (2)³ + 1 

      m²
———————————— =
 8m - 8 + 1 

    m²
———————— =
 8m - 7 

Divide numerator and denominator
by m

     m
————————— =
 8 - 7/m 
 
This is an integer when m = 1 or 7, so
we have solutions (2,7) and (2,1).  (2,1)
is a solution form case 1, so the only new
solution we have is (n,m) = (2,7)

My computer found another solution

(n,m) = (4, 126) 

and I'm not sure how to get it.  Sorry!

What math course is this?  Number theory?

But apparently all the solutions are:

(2m,m), (2k,1), (2,7), and (4, 126)

where m, and k can be any positive integers.

Edwin