Estimate a value with square roots squared without calculating If $n$ is a natural number and $n<(6+\sqrt{29})^2<n+1$, find the value of $n$.

This problem is, of course, very easy to solve with a calculator. But I am looking for a solution that does NOT need to find the value of $(6+\sqrt{29})^2$. Is there a way to do that?
 A: Without a calculator,
$$(6+\sqrt{29})^2=36+12\sqrt{29}+29=65+12\sqrt{29}=125+12(\sqrt{29}-5)=125+{48\over\sqrt{29}+5}$$
Now $5\lt\sqrt{29}\lt6$ implies
$$4\lt{48\over11}\lt{48\over\sqrt{29}+5}\lt{48\over10}\lt5$$
Thus
$$129\lt(6+\sqrt{29})^2\lt130$$
A: Expand the expression to get $36+12\sqrt{29}+29=65+\sqrt{4176}$. Now you only need to find the integer $m$ such that $m<\sqrt{4176}<m+1$, i.e. $m^2<4176<(m+1)^2$.
A: $5<\sqrt{29}<6$, so $0<6-\sqrt{29}<1$, so $0<(6-\sqrt{29})^2<1$.
$(6+\sqrt{29})^2<(6+\sqrt{29})^2+(6-\sqrt{29})^2<(6+\sqrt{29})^2+1$.
But $(6+\sqrt{29})^2+(6-\sqrt{29})^2=36+12\sqrt{29}+29+36-12\sqrt{29}+29=130$.
So $(6+\sqrt{29})^2<130<(6+\sqrt{29})^2+1$.
So $129<(6+\sqrt{29})^2<130$.
A: Use the difference of two squares to get:
$$6+\sqrt{29} = \frac{(6+\sqrt{29})(6-\sqrt{29})}{6-\sqrt{29}} = \frac{7}{6 - \sqrt{29}}$$
and so:
$$n<\frac{49}{(6-\sqrt{29})^2}<n+1 \Rightarrow \frac{1}{n+1} < \frac{(6 - \sqrt{29})^2}{49} < \frac{1}{n}.$$
Adding $49$ times this inequality with the original gives:
$$n + \frac{49}{n+1} < (6 + \sqrt{29})^2 + (6 - \sqrt{29})^2 < n+1 + \frac{49}{n}$$
$$\Rightarrow n + \frac{49}{n+1} < 130 < n+1 + \frac{49}{n}$$
From the right-hand side, the minimum value of $n$ which satisfies the inequality is $129$. But from the left-hand side, $n$ cannot be $130$ or greater. Therefore $n = 129$ which is the unique solution.
