Let $A=\sqrt{13+\sqrt{1}}+\sqrt{13+\sqrt{2}}+\sqrt{13+\sqrt{3}}+\cdots+\sqrt{13+\sqrt{168}}$ and $B=\sqrt{13-\sqrt{1}}+\sqrt{13-\sqrt{2}}+\sqrt{13-\sqrt{3}}+\cdots+\sqrt{13-\sqrt{168}}$.

Evaluate $(\frac{A}{B})^{13}-(\frac{B}{A})^{13}$.

By Calculator, I have $\frac{A}{B}=\sqrt{2}+1$ and $\frac{B}{A}=\sqrt{2}-1$.

But, I don't know how. Has someone any idea about this.

  • 4
    $\begingroup$ Where did you encounter this beauty? $\endgroup$ – Daniel W. Farlow Mar 9 '15 at 6:31
  • $\begingroup$ ^ I want to know that too $\endgroup$ – Hasan Saad Mar 9 '15 at 6:34
  • 1
    $\begingroup$ How did it get 10 votes in 37 minutes? $\endgroup$ – Asaf Karagila Mar 9 '15 at 6:55
  • 1
    $\begingroup$ @crash: Uninformative title with subjective difficulties qualifications? Question without showing any efforts? Those usually garner between two downvotes to two upvotes in a span of days. Certainly not 10 upvotes in less than an hour. $\endgroup$ – Asaf Karagila Mar 9 '15 at 6:57
  • 5
    $\begingroup$ @AsafKaragila Usually is the problem in what you just wrote--the most upvoted questions on MSE are the ones that often get closed in the matter of minutes and garner downvotes themselves. I cannot explain the voting patterns of users here. I imagine what happened to most people, like myself, was they opened a question expecting to find a really easy precalc problem, but then it turned out to be something rather intriguing, hence the upvotes. $\endgroup$ – Daniel W. Farlow Mar 9 '15 at 7:00

This took me some time to solve. Here you go:

First, we find this:

$$\begin{aligned} (\sqrt{13+\sqrt{a}}-\sqrt{13-\sqrt{a}})^2 &=13+\sqrt{a}+13-\sqrt{a}-2\sqrt{13+\sqrt{a}}\sqrt{13-\sqrt{a}}\\ &=2(13-\sqrt{169-a}) \end{aligned}$$



By what we have, we write,


What I used here is the fact that I've summed over all $a$ from $1$ to $168$, and that summing with $\sqrt{169-a}$ is the same as summing with $\sqrt{a}$ in this question.

Now, we have $A=(1+\sqrt{2})B$

Thus, $\frac{A}{B}=1+\sqrt{2}$ and $\frac{B}{A}=\sqrt{2}-1$

We just calculate $(\frac{A}{B})^{13}$ and $(\frac{B}{A})^{13}$ which I believe is okay to be done using calculator. Else, comment so I can edit my answer.

  • $\begingroup$ Nice work. But "which I believe is okay to be done using calculator"--can you show how to reach the integer solution $94642$ manually through algebraic manipulation? $\endgroup$ – Daniel W. Farlow Mar 9 '15 at 7:04
  • $\begingroup$ I'll give it a try. Please wait a bit. $\endgroup$ – Hasan Saad Mar 9 '15 at 7:05
  • $\begingroup$ Cool. Will do. :) Because I think, @AuthawichNarissayaporn correct me if I'm wrong, that computing that is actually part of the problem. $\endgroup$ – Daniel W. Farlow Mar 9 '15 at 7:06
  • $\begingroup$ I'm not really sure I can do it. The best way after my work is to use @math110 solution. He has it pretty nice there to be honest. $\endgroup$ – Hasan Saad Mar 9 '15 at 7:10
  • 1
    $\begingroup$ @crash Sorry, I hit "enter" at the wrong time. Take another look at my message, which shows you how to get to 94642 by algebraic manipulation without using a calculator, as you asked for a couple of days ago. $\endgroup$ – BudgieJane Mar 11 '15 at 17:26

Let $$A=\sum_{n=1}^{168}\sqrt{13+\sqrt{n}},B=\sum_{n=1}^{168}\sqrt{13-\sqrt{n}}$$ since $$\sqrt{2}A=\sum_{n=1}^{168}\sqrt{26+2\sqrt{n}}=\sum_{n=1}^{168}\left(\sqrt{13+\sqrt{169-n}}+\sqrt{13-\sqrt{169-n}}\right)=A+B$$ so we have $x=\dfrac{A}{B}=\sqrt{2}$,then we have $$x=\sqrt{2}+1,\dfrac{1}{x}=\sqrt{2}-1\Longrightarrow x+\dfrac{1}{x}=2\sqrt{2}$$ let $$a_{n}=x^n-x^{-n}$$ use this well know indentity $$a_{n+2}=(x+\dfrac{1}{x})a_{n+1}-a_{n}\Longrightarrow a_{n+2}=2\sqrt{2}a_{n+1}-a_{n}$$ $$a_{1}=2,a_{2}=4\sqrt{2}$$ so $$a_{3}=2\sqrt{2}a_{2}-a_{1}=16-2=14$$ $$a_{4}=2\sqrt{2}a_{3}-a_{2}=28\sqrt{2}-4\sqrt{2}=24\sqrt{2}$$ $$a_{5}=2\sqrt{2}a_{4}-a_{3}=96-14=82$$ $$\cdots$$

  • 2
    $\begingroup$ But, without calculator, we don't know $x=\sqrt{2}+1$. $\endgroup$ – Authawich Narissayaporn Mar 9 '15 at 6:56
  • $\begingroup$ can see @Hasan Saad solution $\endgroup$ – math110 Mar 9 '15 at 6:58
  • $\begingroup$ Thank you for making a solution perfect :) $\endgroup$ – Authawich Narissayaporn Mar 9 '15 at 7:14

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.