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

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Answer 1 · 2015-04-16T15:38:31

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 · 2015-07-10T22:28:57

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

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

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

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