Let me define an operator for two natural numbers as
Now, if , with , find the number of ordered pairs for which is a perfect square of an integer.
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a^b/b^a*b^b/a^a=(ab)^b/(ab)^a=(ab)^b-a. so b-a=are even for perfect square.(suppose, b-a=6 then (ab)^3^2 so if power is 2 then it is square) if b=8 then for all value of a+b=14 there are only 1 perfect square.similarly there are 14 pairs of (a,b)