Let $f (x$) = $\sqrt{−x^2 + 20x + 400} + \sqrt {x^2 − 20x}$. How many elements in the range of $f$ are integers? 
Let $f (x$) = $\sqrt{−x^2 + 20x + 400} + \sqrt {x^2 − 20x}$. How many elements in the range of $f$ are integers?

I first let $y= x^2 -20x +100$. 
Then substitute it in the function ------->  $f(x) = \sqrt{-y+500}+\sqrt{y-100}$.
Which means, $100\leq{y}\leq{500}$. I then tried to find values of $y$ that would make both radicals disappear. I found four values of $y$ that made $f(x)$ integers; $100, 500, 356,$ and $244$. I also checked their discriminants and all of them were greater than $0$, which means each value of $y$ means two solutions for $x$, meaning, there are $8$ elements in the range of $f$ are integers. But the correct answer is $9$ and I can't seem to find the last one.

By C-S $$\sqrt{400+2x-x^2}+\sqrt{x^2-20x}\leq\sqrt{(1+1)(400+2x-x^2+x^2-20x)}=20\sqrt2.$$
Also, since $$\sqrt{a}+\sqrt{b}=\sqrt{a+b+2\sqrt{ab}}\geq\sqrt{a+b},$$ we obtain: $$\sqrt{400+2x-x^2}+\sqrt{x^2-20x}\geq\sqrt{400+2x-x^2+x^2-2x}=20.$$
We got the maximal and the minimal value of $f$ and $f$ is a continuous function.
Can you end it now?
