'Two' many roots

$\Bigg( (\sqrt{2})^1 + (\sqrt{2})^2 \Bigg) \times \Bigg( (\sqrt{2})^3 + (\sqrt{2})^4 \Bigg) \times ... \times \Bigg( (\sqrt{2})^{2017} + (\sqrt{2})^{2018} \Bigg) = 2^k + n$ $(\sqrt{2})^1\Bigg( 1 + \sqrt{2} \Bigg) \times (\sqrt{2})^3\Bigg( 1 + \sqrt{2} \Bigg) \times ... \times (\sqrt{2})^{2017}\Bigg( 1 + \sqrt{2} \Bigg) = 2^k + n$ $(\sqrt{2})^{1} \times (\sqrt{2})^{3} \times (\sqrt{2})^{5} \times ... \times (\sqrt{2})^{2017} \times \Bigg( 1 + \sqrt{2} \Bigg)^{1009} = 2^k + n$ $(\sqrt{2})^{1 + 3 + 5 + ... + 2017} \times \Bigg( 1 + \sqrt{2} \Bigg)^{1009} = 2^k + n$ Since $\displaystyle \sum_{n=1}^{k}2n-1 = k^2$ , $(\sqrt{2})^{(1009^2)} \times \Bigg( 1 + \sqrt{2} \Bigg)^{1009} = 2^k + n$ $(\sqrt{2})^{(1018081)} \times \Bigg( 1 + \sqrt{2} \Bigg)^{1009} = 2^k + n$ $(\sqrt{2})^{(1018080)} \times (\sqrt{2})^{(1)} \times \Bigg( 1 + \sqrt{2} \Bigg)^{1009} = 2^k + n$ $2^{509040} \times \sqrt{2} \times \Bigg( 1 + \sqrt{2} \Bigg)^{1009} = 2^k + n$ Now let us consider the binomial expression of $(1 + \sqrt{2})^{1009}$ .

Let $(1 + \sqrt{2})^{1009} = a + b\sqrt{2}$ . $2^{509040} \times \sqrt{2} \times (a + b\sqrt{2}) = 2^k + n$ $2^{509040} \times \Bigg(a\sqrt{2} + 2b\Bigg) = 2^k + n$ As $a$ will always be an odd integer, the bit in brackets will be irrational so will have no integer powers of $2$ as factors, meaning that $\boxed{509040}$ is the greatest power of $2$ in the left-hand side, so it is the greatest possible value of $k$ . Therefore, $k+1=\boxed{509041}$