<s>[INST] 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.
 [/INST] Let $z=y-300$.  You want to solve $\sqrt{200-z}+\sqrt{200+z}=N,$ where $N$ is an integer.
Square both sides:  $400+2\sqrt{200^2-z^2}=N^2$ or $\sqrt{200^2-z^2}=\dfrac{N^2-400}2,$
which means $0\le\dfrac{N^2-400}2\le200$ or $400\le N^2\le800.$
Note we must have $N\ge0$.  Can you take it from here?
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