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Using 5000-digit precision in PARI/GP, I discovered that the fractional part of $(\sqrt{2}+\sqrt{3})^{2008}$ is extremely small, less than $10^{-999}$. Is there a simple explanation for this fact ?

This looks like a Pisot number issue (similar questions have already been studied on MSE, see for example Why is $(2+\sqrt{3})^{50}$ so close to an integer?), but it’s a more complicated situation.

Related : Show that $(\sqrt{2} + \sqrt{3})^{2009}$ is rounded to an even number.

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  • $\begingroup$ How is this different from the first question you linked to? $(\sqrt{2}+\sqrt{3})^{2008} = (5 + 2\sqrt{6})^{1004}$ and $5 + 2\sqrt{6}$ is a Pisot integer... $\endgroup$
    – A.P.
    Apr 27, 2015 at 7:57
  • $\begingroup$ Also, could you please accept Daniel Fischer's answer if it satisfied you, or could you tell us why it didn't, so that we can improve on it? $\endgroup$
    – A.P.
    Apr 27, 2015 at 7:58
  • $\begingroup$ @A.P. My question is different because it is about numbers in a biquadratic field while the linked question is about a quadratic field (even though your argument reduces the former to the latter). In the same vein, Daniel Fischer’s answer involves binomial expansion while the linked question only needs conjugation in a quadratic field. $\endgroup$ Apr 27, 2015 at 12:49
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    $\begingroup$ @A.P. For some reason, I forgot to accept Daniel Fischer’s answer, thank you for reminding me. On the other hand, I find your argument better and simpler than Daniel Fischer’s, so if you turn your comment into an answer I’ll accept it. But you seem to consider it unworthy of constituting a real answer … $\endgroup$ Apr 27, 2015 at 12:50
  • $\begingroup$ Done. I posted it as a comment because I thought you were asking for more... $\endgroup$
    – A.P.
    Apr 27, 2015 at 18:25

2 Answers 2

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That is because

$$(\sqrt{3}+\sqrt{2})^{2m} + (\sqrt{3}-\sqrt{2})^{2m}$$

is an integer, namely

$$\begin{align} (\sqrt{3}+\sqrt{2})^{2m} + (\sqrt{3}-\sqrt{2})^{2m} = 2\sum_{k=0}^{m} \binom{2m}{2k} 3^{m-k}2^k, \end{align}$$

and $(\sqrt{3}-\sqrt{2})^{2m} <\frac{1}{3^{2m}}$.

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    $\begingroup$ Yes, actually, $\sqrt{3}-\sqrt{2} < \frac{1}{3}$, which would yield a better bound, $\frac{1}{9^{1004}}$. That's in the right ballpark, but of course a more exact estimate gives more exact results. $\endgroup$ Jun 9, 2014 at 19:47
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    $\begingroup$ @EwanDelanoy calling the integer with exponents $2m$ by $x_m,$ we have $x_0 = 2, x_1 = 10, x_2 = 98,$ and $x_{m+2} = 10 x_{m+1}- x_m.$ $\endgroup$
    – Will Jagy
    Jun 9, 2014 at 20:06
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Your intuition is correct: $$ (\sqrt{2} + \sqrt{3})^{2008} = (5 + 2 \sqrt{6})^{1004} $$ and $5 + 2\sqrt{6}$ is a Pisot integer. Further, Newton's identities imply that $$ (5 + 2\sqrt{6})^n + (5 - 2\sqrt{6})^n \in \Bbb{Z} $$ for every positive integer $n$, thus the distance between $(5 + 2\sqrt{6})^{1004}$ and the closest integer is at most $$ (5 - 2\sqrt{6})^{1004} \approx 2.6743 \cdot 10^{-1000} $$

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