SOLUTION: Solve : x! × y! = 120 × (xy)!

There are two solutions.



One solution is     x= 5, y= 0.   

Other solution is   x= 0, y= 5.   



Let's prove first that  x= 5, y= 0 is the solution.



Indeed, then  x! = 5! = 120;

              y! = 0! = 1   ( by the definition of 0! )

        thus left side is  x! * y! = 120 * 1 = 120.



Next, for the right side  (xy)! = (5*0)! = 0! = 1 ( again, by the definition of 0! ),

and thus left side, 120, is equal to the right side, which is 120, too.



For other solution, x= 0, y= 5,  the proof is symmetric.

Ikleyn has found two solutions, (x,y) = (5,0) and (x,y) = (0,5) 
but has failed to show that there are no other solutions.  But it is
necessary to show that there are no other solutions, not just to
find two solutions and verify that they are both solutions.



Those will be the only solutions if and only if

, for  and 

Assume 

If x = 0, then , except for the solution (x,y) = (0,5)

If x = 1, then 

If  then 

The numerator of that last fraction has xy-x indicated factors and the
denominator has y factors.

Since  and ,


Subtract 1 from both sides

Divide both sides by x-1









This proves the numerator has at least as many indicated factors as the
denominator.  The first x factors of the denominator are, respectively, less 
than or equal to the first x factors of the numerator.  The remaining indicated
factors of the numerator, if any, are even greater. Therefore, the numerator of
the fraction  is greater than the denominator, and the fraction 
is greater than or equal to 1, and therefore not equal to .

The case  is proved by symmetry.

Therefore the 2 solutions Ikleyn found above are the ONLY solutions.

Edwin