Question

For how many real values of x is =" \sqrt{120-\sqrt{x}} " alt=" \sqrt{120-\sqrt{x}} " align="absmiddle" class="latex-formula"> an integer?

Answer 1

A good place to start is to set $\sqrt{x}$ to y. That would mean we are looking for $\sqrt{120-y}$ to be an integer. Clearly, $y\leq 120$, because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since $\sqrt{x}$ is a radical, it only outputs values from $[0,\infty]$, which means y is on the closed interval: $[0,120]$.

With that, we don't really have to consider y anymore, since we know the interval that $\sqrt{x}$ is on.

Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval $[0,120]$, which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:

$\sqrt{120-\sqrt{x}}=k \implies \\ \sqrt{x}=k^2-120 \implies\\ x=(k^2-120)^2$

Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.