<s>[INST] 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?
 [/INST] 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.
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