Vtyjkyuk

$sqrt3=1.b_1b_2...$is the binary representation of $sqrt3$.

i.e. $sqrt3=1+dfracb_12^1+dfracb_22^2+...$

Prove that at least one of the digits $b_n,b_n+1,...,b_2n$ is 1.

my attempt:

Square both sides: $$left(1+sum_i=1^inftyfracb_i2^iright)^2=3$$
Expand and subtract $1$ from both sides
$$2sum_i=1^inftyfracb_i2^i+sum_i=1^inftyfracb_i^22^2i=2$$
divide both side by 2
$$sum_i=1^inftyfracb_i2^i+sum_i=1^inftyfracb_i^22^2i+1=1$$
Tidy up
$$sum_i=1^inftyfrac12^ileft(b_i+fracb_i^22^i+1right)=1$$
Lemma: $displaystylesum_i=m^inftyfrac12^ileft(1+frac12^i+1right)<frac12^m-2$

Proof: multiply both side by $2^m-1$$$sum_i=1^inftyfrac12^i+frac12^m+1+i<2$$which is trivial for any positive integer m.

Proceed the proof by contradiction: suppose $b_n,b_n+1,...b_2n$ are all equal to $0$. Then the sum $sum_i=1^n-1frac12^ileft(b_i+fracb_i^22^i+1right)$ is in the form of $1-fracx2^2n-1$.

But $$sum_i=2n+1^inftyfrac12^ileft(b_i+fracb_i^22^i+1right)<frac12^2n-1$$by lemma.

Q.E.D.

Is this proof correct?
I am also happy for an alternative (and hopefully shorter) solution.

The flaw in your proof was pointed out in the comments, so let me show you my approach

Assume $sqrt 3$ has all digits $b_i=0$ for $iinn,...,2n$ (these are $n+1$ digits). Write

$$sqrt 3cdot 2^n-1=N + epsilon= b_1cdots b_n-1.underbrace0cdots0_n+1b_2n+1cdots$$

where $N=b_1cdots b_n-1inBbb N$ is the integer part, and $epsilon=0.b_nb_n+1cdotsin(0,1)$ is the fractional part (which is stricly positive because of the irrationality of $sqrt3$). We have $N< 1.8cdot 2^n-1$ (with $1.8$ being a rough upper bound for $sqrt 3$). Further, $epsilon$ must have zeros as the first $n+1$ digits $b_n,...,b_2n$. So the largest possible value is

$$epsilon le 0.underbrace0dots0_n+1overline 1=0.underbrace0cdots 0_n1=frac12^n+1$$

Now observe that

$$underbrace3cdot 2^2n-2_textinteger=(sqrt 3cdot2^n-1)^2=N^2+2Nepsilon+epsilon^2.$$

The left side is an integer, and so is $N^2$. Note that $0<2Nepsilon< 0.9$. And finally, since $epsilon^2<1/2^2n+2<0.1$, we cannot close the gap to make the right side a whole integer too.