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

Apr 16, 2015

I think this is pretty clear:

For all $c$ satisfying these:
$\sqrt{60} < c < \sqrt{70} \mathmr{and} - \sqrt{70} > c > - \sqrt{60}$
${c}^{2}$ will be between $60$ and $70$.

Since there is no natural number between $\sqrt{60}$ and $\sqrt{70}$ there won't be any $c \in \mathbb{N}$

Jul 10, 2015

${\left\lceil \sqrt{60} \right\rceil}^{2} = {\left\lfloor \sqrt{70} \right\rfloor}^{2} = {8}^{2} = 64$

#### Explanation:

$7 < \sqrt{60} < 8 < \sqrt{70} < 9$

So $\left\lceil \sqrt{60} \right\rceil = 8 = \left\lfloor \sqrt{70} \right\rfloor$

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 $n \ge 22$, we find

$64 < \frac{64 {\left(n + 1\right)}^{2}}{n} ^ 2 < 70$