Find all possible
nonzero integer solutions to this equation:
13 1996 Z
 +  = 
M^{2} N^{2} 1997
Prove that no further solution to the above equation is possible.
If (M,N,Z) is a solution so are (M,N,Z) (M,N,Z) (M,N,Z) so it is sufficient to consider M>0, N>0 (Z by inspection must be positive.)
It is also clear that the solution set must be finite because the LHS gets smaller as the RHS gets bigger as M,N,Z increase.
There are two solutions if M=N since the equation becomes
2009*1997/M^2=Z
1997 is prime but 49 is a factor of 2009 so we have
(1,1,4011973)
(7,7,81877)
Five further solutions come from letting N=2M since the equation becomes
2048*1997/(4M^2)=1997*512/(M^2)=Z and 512=2^9 so has five perfect square factors. The solutions are
(1,2,1022464)
(2,4,255616)
(4,8,63904)
(8,16,15976)
(16,32,3994)
I don't know if there are more than these 7 (or 28 if you allow negatives.)

Posted by Jer
on 20161009 16:59:52 