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Assume that $a_0=-2$, $b_0=1$, and that, for every $n\ge0$, $$a_{n+1}=a_n+b_n+\sqrt{a^2_n+b^2_n} \qquad b_{n+1}=a_n+b_n-\sqrt{a^2_n+b^2_n}$$

How to find $a_{2012}$?

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    Why didn't Math.StackExchange start 2000 years earlier? It would make life much more [easier...](http://math.stackexchange.com/search?q=2012)2012-07-19
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    Hint: What is $a_{n+1}+b_{n+1}$, and what is $a_{n+1}^2+b_{n+1}^2$?2012-07-19
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    Got something from an answer below?2012-07-25
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    Bis repetita placent.2012-07-29
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    @did I added them and multiplied them ,but what is the next step ?2012-07-29
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    Sorry but this is still very vague. If you are referring to my answer, precisely which steps of the plan I suggest, did you perform?2012-07-29
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    @did not your answer but i think in your answer its step 3 .2012-07-29
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    OK. What about steps 1-2-4-5-6-7? What is the first one you have a problem with?2012-07-29
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    @did I think i have no problem with 1,2 and 3 ,step 4 is the first one that i have a problem with.2012-07-29
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    Step 4 suggests to compute s0=a0+b0s_0=a_0+b_0 and p0=a0b0p_0=a_0b_0 and you **KNOW** what a0a_0 and b0b_0 are...2012-07-29
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    @did is that what the other answer also suggested ?2012-07-29
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    How is this relevant? You were supposed to indicate the first step of the plan I suggested, that you have a problem with. You indicated Step 4, which I find unbelievable, so which step is it? 4? 5? A later one?2012-07-29
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    @did i understand step 4 ,it is 52012-07-29
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    Well, well, well... So your problem is as follows: you are given the sum s=a+b and the product p=ab of two unknown numbers a and b, and you want to compute a and b. Any idea?2012-07-29
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    @did This is what i am trying to ask about , i have the sum and the product but how to do the next step ?2012-07-29
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    Consider this: what are the roots of $x^2-sx+p=0$? And what are the roots of $x^2-(a+b)x+ab=0$?2012-07-29
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    $a_n$ and $b_n$ the roots of the quadratic $\lambda^2 + 2^n\lambda - 2^{n+1} = 0$ ? So they are $-2^{n-1} \pm \sqrt{2^{2(n-1)} + 2^{n+1}}$. Which is $a_n$ Since $a_{n+1} - b_{n+1} = 2\sqrt{a_n^2 + b_n^2} > 0$, it follows $a_n > b_n$ for $n\geq 1$, so $a_n = -2^{n-1} + \sqrt{2^{2(n-1)} + 2^{n+1}}$2012-07-29
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    You are welcome. As you know, the next step of the standard procedure is to *accept* an answer.2012-08-11
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    @did,i have accepted your "7-steps plan".2012-09-20
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    Cool. $ $ $ $ $ $2012-09-20

1 Answers 1

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Here is a 7-steps plan:

  1. Stop asking questions with no indication whatsoever about which similar problems you can solve, what you tried to solve the present one, and why you think your attempts failed.

  2. Stop ignoring comments asking you to add the pieces of information mentioned in 1.

  3. Define $s_n=a_n+b_n$ and $p_n=a_nb_n$ and, for every $n\geqslant0$, express $s_{n+1}$ and $p_{n+1}$ in terms of $s_n$ and $p_n$.

  4. Compute $s_0$ and $p_0$.

  5. For every $n\geqslant1$, express $a_n$ and $b_n$ in terms of $s_n$ and $p_n$.

  6. Conclude.

  7. Memorize 1. and 2. for your next questions on the site.