2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ This is an irrational number. How many decimals ? $\endgroup$ – Yves Daoust Jan 27 at 12:19
4
$\begingroup$

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

$$129\lt(6+\sqrt{29})^2\lt130$$

$\endgroup$
5
  • $\begingroup$ Nice proof (+1) $\endgroup$ – Angelo Jan 27 at 12:25
  • $\begingroup$ Why make the denominator $\sqrt{29}+5$? Is +4 or +3 fine? $\endgroup$ – Cyh1368 Jan 27 at 13:23
  • 1
    $\begingroup$ @Cyh1368, $\sqrt{29}+5$ is the "conjugate" to $\sqrt{29}-5$, i.e., $(\sqrt{25}+5)(\sqrt{29}-5)=29-25=4$. $\endgroup$ – Barry Cipra Jan 27 at 14:41
  • $\begingroup$ @BarryCipra But what if we used $\sqrt{29}-4$ in the first place? Then can't we just ues $\sqrt{29}+4$ as the conjugate? $\endgroup$ – Cyh1368 Jan 28 at 2:22
  • $\begingroup$ @Cyh1368, ah, I see what you're getting at: How did I know to turn $65+12\sqrt{29}$ into $125+12(\sqrt{29}-5)$ rather than $113+12(\sqrt{29}-4)$? The idea is to use a $12(\sqrt{29}-a)$ that makes the numerator, $12(29-a^2)$, as small as possible. $\endgroup$ – Barry Cipra Jan 28 at 11:27
3
$\begingroup$

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. $m^2<4176<(m+1)^2$.

$\endgroup$
3
  • $\begingroup$ It's a simple solution, but still need some time to figure out $\sqrt{4176}$ $\endgroup$ – Cyh1368 Jan 27 at 13:01
  • $\begingroup$ @Cyh1368: Yes, you are right -- I prefer the other solutions to mine. (But as every programmer knows, $64^2=4096$.) $\endgroup$ – TonyK Jan 27 at 13:18
  • $\begingroup$ (+1), and $65^2 = (6\times 7)\times 100+25=4225$ is a well-known trick too. $\endgroup$ – Neat Math Jan 27 at 14:12
2
$\begingroup$

$5<\sqrt{29}<6$, 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 $129<(6+\sqrt{29})^2<130$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Note that solving this inequality by expanding and using the quadratic formula would be horrendous. $\endgroup$ – Toby Mak Jan 27 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.