MHB Prove a_n=⌊2^n√2⌋ contains infinitely many composite numbers

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Prove, that the sequence:

$a_n = \left\lfloor{2^n\sqrt{2}}\right\rfloor,\;\;\;\;n = 1,2,3,..$- contains infinitely many composite numbers.
 
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Suggested solution:
Suppose that the sequence ${a_n}$ contains only finitely many composite numbers, i.e. there exists an $N$ such that for all $n>N$, $a_n$ is odd.

Consider the binary representation of $a_n$ ($n>N$). The last digit of $a_n$ must be $1$, which in turn implies, that the fractional part of binary representation of $\sqrt{2}$ has all its digits equal to $1$ after the $N$´th position. This contradicts the irrationality of $\sqrt{2}$.
 
lfdahl said:
Suggested solution:
Suppose that the sequence ${a_n}$ contains only finitely many composite numbers, i.e. there exists an $N$ such that for all $n>N$, $a_n$ is odd.

Consider the binary representation of $a_n$ ($n>N$). The last digit of $a_n$ must be $1$, which in turn implies, that the fractional part of binary representation of $\sqrt{2}$ has all its digits equal to $1$ after the $N$´th position. This contradicts the irrationality of $\sqrt{2}$.

Hats off to you
 
kaliprasad said:
Hats off to you

The solution is not mine :o
 
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