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The question is how to find first 10 digits after decimal point in the number $(1+\sqrt{3})^{2015}$.

I keep running into this kind of problems in a context of symmetric polynomials.

Glinka
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2 Answers2

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Hint: (i) The number $(1+\sqrt{3})^{2015}+(1-\sqrt{3})^{2015}$ is an integer. We can see this by imagining expanding using the Binomial Theorem. Terms involving odd powers of $\sqrt{3}$ cancel.

(ii) The number $(1-\sqrt{3})^{2015}$ is extremely small in absolute value, and negative. So $(1+\sqrt{3})^{2015}$ is nearly an integer.

Arthur
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André Nicolas
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Another way to see that $u_n=(1+\sqrt{2})^n+(1-\sqrt{2})^n$ is an integer is to exhibit a simple three-term recursion for it. $$ u_{n+1} = 2u_n+u_{n-1} $$

GEdgar
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