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

The answer was 1154.

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.