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

[lfdahl]

Prove, that the sequence:

$a_n = \left\lfloor{2^n\sqrt{2}}\right\rfloor,\;\;\;\;n = 1,2,3,..$- contains infinitely many composite numbers.

 

[lfdahl]

Suggested solution:

Spoiler

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 $\sqrt{2}$.

 

[kaliprasad]

Suggested solution:

Spoiler

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 $\sqrt{2}$.

Spoiler

Hats off to you

 

[lfdahl]

Spoiler

Hats off to you

The solution is not mine