The question is
$ f(n) = \sqrt{{100}+\sqrt{n}} + \sqrt{{100}-\sqrt{n}} $
What is the minimum value of n for which $f(n)$ is an integer. $n$ is a natural number.
I made an attempt to solve it, and although I can't find anything wrong with my method, I still have the wrong answer.
first, I set the function equal to some integer $p$ and then squared both sides to get
$ 200 + 2\sqrt{100^2-n} = p^2$ –> 'equation A'
note: $100^2 – n$ must be a perfect square
I then factored out the 2. But because it is equal to $p^2$, a square, and 2 is prime, $100 + \sqrt{100^2-n}$ must also have a 2 to be factored out, 100 is already a multiple of 2, so, the radical must also be a multiple of 2, giving us
$\sqrt{100^2-n} = 2q$ –> so, $100^2-n = 4q^2$ for some integer q
since $n > 0$, $-50 < q < 50$
if we substitute $4q^2$ into equation A with the 2 factored out, we get
$2(100 \pm 2q) = p^2$
factoring out a 2 again, we get
$2^2(50 \pm q) = p^2$
so, $2\sqrt{50 \pm q} = p$ meaning that $50 \pm q$ is a perfect square.
using the earlier inequality for $q$ we get
$0 < 50 \pm q < 100$
solving for all squares in this range, $q$ could be $49, 46, 41, 34, 25, 14, 1$ or $31$ (I think we can ignore the $\pm$ sign as the q will be squared)
using $100^2 – n = 4q^2$ and rearranging for $n$, we can find the minimum value of $n$ by using the maximum value of $q$. I got $396$ as my answer but the answer is $6156$. Please help me find what went wrong in my method and why it gave the wrong answer.
Best Answer
By squaring both sides of the equation we get $p^{2}=200+2\sqrt{100^{2}-n}<400$ (putting $n=0$) so that $p<20$.
By rearranging terms we get $$\left(\frac{p^{2}-200}{2}\right)^{2}=100^{2}-n$$that implies that $p$ should be even.
The smallest $n$ is obtained with the biggest $p$ (an even number smaller than $20$). For $p=18$ we get $n=6156$.