4meonly wrote: Guys, where is my mistake?

х can be as 1 as 9, and they are both odd!

Answer E.

Where is my mistake?

You didn't make a mistake! Your solution is perfect, and the answer should be E. As 4meonly pointed out, x can be 9:

x^x = 9^9 = (3^2)^9 = (3^9)^2

So if x = 9, y would then be 3^9, which is certainly an integer. Indeed, there are many other possible values of x- x could be any square of an odd integer.

dalwow wrote:Ian-

Is there a faster way to approach this problem? It doesn't seem like we would be expected to work it out for all the digits. Additionally, just multiplying it out for 9^9 and finding the square root of that would take considerable time. What are the <2 minute approaches?

Well, no, there's no need to even choose a number here. The answer is clearly C or E. Using both statements, we only know

x^x = y^2

and that x and y are odd integers. That is, the question is really asking, if x is odd, how can x^x be a perfect square? Is x^x only a perfect square when x = 1? Well, no. If I raise any odd perfect square to any positive integer power x, I'll get another perfect square, so I just need to make x = p^2 for any odd number p. There are thus infinitely many solutions for x.

Algebraically, if x = p^2, then x^x = (p^2)^x = (p^x)^2. So if x = p^2, y will be p^x or -p^x.