Question

How do you find the perfect square between 60 and 70?

Answer 1

I think this is pretty clear:

For all `c` satisfying these:
`sqrt(60) < c < sqrt(70) and -sqrt(70) > c > -sqrt(60)`
`c^2` will be between `60` and `70`.

Since there is no natural number between `sqrt(60)` and `sqrt(70)` there won't be any `c in NN`

Answer 2

`ceil(sqrt(60))^2 = floor(sqrt(70))^2 = 8^2 = 64`

Explanation:

`7 < sqrt(60) < 8 < sqrt(70) < 9`

So `ceil(sqrt(60)) = 8 = floor(sqrt(70))`

and `60 < 8^2 = 64 < 70`

`64` is the only perfect square integer between `60` and `70`, but there are an infinite number of square rational numbers between `60` and `70`.

In fact for any integer `n >= 22`, we find

`64 < (64(n+1)^2) / n^2 < 70`