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define sequence $$S_{n}=[2^n\cdot \sqrt{2}],n\in N$$

show that

$\{S_{n}\}$contains infinitely many composite numbers

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where $[x]$ is the largest integer not greater than $x$

my try: since

$$S_{4}=[16\sqrt{2}]=22$$ is composite number.

$$S_{5}=[32\sqrt{2}]=45$$ is also composite number

$$S_{6}=[64\sqrt{2}]=90$$ is also composite number.

$$S_{8}=[256\sqrt{2}]=362$$

is also composite number.

and so on.

maybe this problem use Pell equation to solve it.But I can’t .Thank you

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Let $\{x\}$ be $x-[x]$. Note that if $\{2^n\sqrt2\}<0.5$, then

$$[2^{n+1}\sqrt2]=[2(2^n\sqrt2)]=[2([2^n\sqrt2]+\{2^n\sqrt2\})]=[2[2^n\sqrt2]+2\{2^n\sqrt2\}]=2[2^n\sqrt2]$$

and it’s composite.

And this happens infinitely many times, else there would be some $N$ such that for all $n\ge N$ we would have $\{2^n\sqrt2\}\ge0.5$. But then

$$\{2^n\sqrt2\}-\{2^{n+1}\sqrt2\}=\{2^n\sqrt2\}-(2\{2^n\sqrt2\}-1)=1-\{2^n\sqrt2\},$$

so $\{2^{n+1}\sqrt2\}<\{2^n\sqrt2\}$ and the difference is increasing as $n$ increases, which is not possible.

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