# How prove this sequence $S_{n}=,n\in N$ contains infinitely many composite numbers

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

show that

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

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

#### Solutions Collecting From Web of "How prove this sequence $S_{n}=,n\in N$ contains infinitely many composite numbers"

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.