We see that n is in the set of natural numbers (1,2,3,4,5,...), so of course m should be in the natural numbers. However, we see that 2n^2, when square rooted, is [tex]n\sqrt{2}[/tex]. With n in the natural numbers, the square root of m, or o, can never be a natural number but is always an irrational number. Therefore, this equation has no solutions.
