third option is the correct answer....
No; we have 
. Substituting these into the DE gives
which reduces to 
, true only for 
.
A good place to start is to set 
to y. That would mean we are looking for 
to be an integer. Clearly, 
, 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 
is a radical, it only outputs values from ![[0,\infty]](https://tex.z-dn.net/?f=%20%5B0%2C%5Cinfty%5D%20)
, which means y is on the closed interval: ![[0,120]](https://tex.z-dn.net/?f=%20%5B0%2C120%5D%20)
.
With that, we don't really have to consider y anymore, since we know the interval that is on.
Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval , 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:
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.
