Find three squares such that each minus the product of the three gives a square.
Classical Rules: Let a "square" be any number that is the square of a rational number.
(In reply to Partial Solution
I first tried to see if there were any solutions where the three squares were equal other than (1,1,1)
If each is rational (say a/b) then
(a/b)^2 - (a/b)^6 = (c/d)^2 where c/d is rational.
This simplifies to (b^4-a^4) = [((b^3)*c)/(ad)]^2 in which case we know from Fermat's Last Theorum that the difference of powers to the fourth cannot equal a square among integers, and since the difference is a non-zero integer (remember I am excluding (1,1,1) its square root cannot be rational without it being a square, which it isn't. Therefore at least there is no other solution with three equal squares.
Edited on April 20, 2006, 10:51 pm
Posted by Eric
on 2006-04-20 22:19:40