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# Determine the integer closest to $(1+\sqrt 2)^8$

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I need help determining which integer is closest to $(1+\sqrt 2)^8$

Thanks!

May 7, 2022

#1
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The integer closest to (1 + sqrt(2))^8 is 1648.

May 7, 2022
#2
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That was incorrect. Are you sure?

Guest May 7, 2022
#3
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May 7, 2022
#4
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Observation: $$(1 + \sqrt 2)^8 + (\sqrt 2-1)^8$$ is an integer, and $$0 < (\sqrt 2- 1)^8 < \dfrac12$$. That integer is the required integer.

Using binomial theorem,

$$\quad(1 + \sqrt 2)^8 + (\sqrt 2-1)^8\\ =\displaystyle\sum_{k = 0}^8 \binom{8}k ((\sqrt 2)^k + (-\sqrt 2)^k)\\ =\displaystyle\binom{8}0 (2) + \binom82 (2(\sqrt 2)^2) + \binom84 (2(\sqrt 2)^4) + \binom86 (2(\sqrt 2)^6) + \binom88 (2(\sqrt 2)^8)\\ = 1154$$

Note that $$\sqrt 2 - 1 \approx 0.414$$, so $$0 < (\sqrt 2 - 1)^8 < \sqrt 2 - 1 < \dfrac12$$. That means $$(1 + \sqrt 2)^8$$ plus a real number way smaller than 1/2 gives 1154.

Therefore, the nearest integer to $$(1 + \sqrt 2)^8$$ is 1154.

Remarks: The error is $$\left|(1 + \sqrt2)^8 - 1154\right| = (\sqrt 2 - 1)^8 \approx 8.665517772 \times 10^{-4}$$, which is really really small.

May 9, 2022