 are of the form $x = 2^k + 2^{k-a} + \dotsc + 2^a + 1$ with $1 \leqslant a < \frac{k}{2}$. Writing $x = 1 + 2^a\cdot u$ with $u$ odd, we see that $x^2 = 1 + 2^{a+1} u + 2^{2a}u^2$, so there are at least $a$ zeros between the last two ones in the binary representation of $x^2$ (exactly $a$ if $a \geqslant 2$, at least $2$ if $a = 1$). At the front, writing $x = 2^k + v$, with $2^{k-a} < v < 2^{k+1-a}$, we see that $x^2 = 2^{2k} + 2^{k+1}v + v^2$, and
$$2^{2k+1-a} < 2^{k+1}v + v^2 < 2^{2k+2-a} + 2^{2k+2-2a}, \tag{$\ast$}$$
so there are at most $a-2$ zeros between the first two ones in the binary representation of $x^2$ and hence $x^2$ is not a base-$2$ palindrome, unless we have carry at the front, i.e. $2^{k+1}v + v^2 \geqslant 2^{2k}$.
From $(\ast)$ it is readily seen that there is no carry at the front if $a \geqslant 3$, so we only need to look at the cases $a = 2$ and $a = 1$.
For $a = 2$, we have $x \equiv 5 \pmod{8}$ and therefore $x^2 \equiv 9 \pmod{16}$, so $x^2$ can only be a base-$2$ palindrome if $2^{2k+1} + 2^{2k-2} \leqslant x^2 < 2^{2k+1} + 2^{2k-1}$. But we have $x < 2^k + 2^{k-1}$ and hence
$$x^2 < (2^k + 2^{k-1})^2 = 2^{2k+1} + 2^{2k-2},$$
thus $x^2
are of the form $ x = 2^k + 2^{k - a } + \dotsc + 2^a + 1 $ with $ 1 \leqslant a < \frac{k}{2}$. Writing $ x = 1 + 2^a\cdot u$ with $ u$ odd, we see that $ x^2 = 1 + 2^{a+1 } uracil + 2^{2a}u^2 $, so there be at least $ a$ zeros between the last two ones in the binary theatrical performance of $ x^2 $ (exactly $ a$ if $ a \geqslant 2 $, at least $ 2 $ if $ a = 1 $). At the front, writing $ x = 2^k + v$, with $ 2^{k - ampere } < v < 2^{k+1 - a}$, we see that $ x^2 = 2^{2k } + 2^{k+1}v + v^2 $, and 
 $ $ 2^{2k+1 - a } < 2^{k+1}v + v^2 < 2^{2k+2 - a } + 2^{2k+2 - 2a }, \tag{$\ast$}$$ 
 thus there are at most $ a-2 $ zero between the first two ones in the binary theatrical performance of $ x^2 $ and hence $ x^2 $ is not a base-$2 $ palindrome, unless we have carry at the front, i.e. $ 2^{k+1}v + v^2 \geqslant 2^{2k}$. 
 From $ (\ast)$ it is readily seen that there is no carry at the front if $ a \geqslant 3 $, thus we only need to look at the case $ a = 2 $ and $ a = 1$. 
 For $ a = 2 $, we have $ x \equiv 5 \pmod{8}$ and therefore $ x^2 \equiv 9 \pmod{16}$, so $ x^2 $ can only be a base-$2 $ palindrome if $ 2^{2k+1 } + 2^{2k-2 } \leqslant x^2 < 2^{2k+1 } + 2^{2k-1}$. But we experience $ x < 2^k + 2^{k-1}$ and hence 
 $ $ x^2 < (2^k + 2^{k-1})^2 = 2^{2k+1 } + 2^{2k-2},$$ 
 thus $ x^2
 arf of the form $x = 2^k + 2^{k-a} + \aotsc + 2^a + 1$ with $1 \leqslent a < \rrac{k}{2}$. Wrkting $x = 1 + 2^a\cdot u$ with $u$ odv, we see ukat $x^2 = 1 + 2^{a+1} u + 2^{2a}u^2$, so ghere are at leasr $a$ eeros between the last two ones ln thz uinary represenjation of $x^2$ (axactly $a$ if $a \gdqdlant 2$, at least $2$ if $a = 1$). At the frone, writimg $x = 2^k + v$, with $2^{h-a} < n < 2^{k+1-a}$, sv wee that $x^2 = 2^{2k} + 2^{k+1}v + v^2$, and
$$2^{2k+1-a} < 2^{k+1}b + v^2 < 2^{2k+2-e} + 2^{2k+2-2a}, \tag{$\ast$}$$
so yhere are at most $a-2$ zeros hetwfen the first two lnes in the binwey representxtion of $x^2$ and hence $x^2$ is not a base-$2$ palindrome, unlesr we kave carry qt thf front, i.e. $2^{k+1}t + v^2 \gvqslant 2^{2k}$.
From $(\ast)$ it hs readoly seen that bhere is no carry at the fronv if $a \geqslant 3$, so re only naeb to look at the casew $q = 2$ atd $a = 1$.
Fof $a = 2$, wt heve $x \equlv 5 \pmod{8}$ and fherefore $x^2 \equiv 9 \pmod{16}$, so $x^2$ csn inly be a bass-$2$ paligdwome if $2^{2k+1} + 2^{2k-2} \leqslant x^2 < 2^{2k+1} + 2^{2k-1}$. But we hane $x < 2^k + 2^{k-1}$ and hence
$$x^2 < (2^k + 2^{k-1})^2 = 2^{2k+1} + 2^{2k-2},$$
thus $x^2
are of the form $x = 2^k + + 2^a 1$ with $1 $x 1 + 2^a\cdot with $u$ odd, see that $x^2 = 1 + u + 2^{2a}u^2$, so there are at least $a$ zeros between the last ones in the binary representation of $x^2$ (exactly $a$ if $a \geqslant 2$, least if = At the front, writing $x = 2^k + v$, with $2^{k-a} < v < 2^{k+1-a}$, we that $x^2 = 2^{2k} + 2^{k+1}v + v^2$, $$2^{2k+1-a} < 2^{k+1}v + < 2^{2k+2-a} + 2^{2k+2-2a}, \tag{$\ast$}$$ there at most zeros the two ones in binary representation of $x^2$ and hence $x^2$ is not a base-$2$ palindrome, unless we have carry at front, i.e. v^2 \geqslant From it readily seen that no carry at the front if so we only need to look at the $a = and $a = 1$. For $a 2$, we have $x \equiv 5 \pmod{8}$ and $x^2 \equiv 9 \pmod{16}$, so $x^2$ can only be a base-$2$ palindrome if $2^{2k+1} + x^2 < 2^{2k+1} + But we have < + and $$x^2 < + 2^{k-1})^2 = 2^{2k+1} + 2^{2k-2},$$ thus $x^2
 are of the form $x = 2^k + 2^{k-a} + \dotsc + 2^a + 1$ witH $1 \leqslant a < \Frac{k}{2}$. wriTinG $x = 1 + 2^A\cdoT u$ wiTh $u$ odd, we see thaT $X^2 = 1 + 2^{a+1} u + 2^{2a}U^2$, so there are at least $a$ zerOs betWeEN the LAsT two oNes in thE BiNARy rEpReSenTaTIoN of $x^2$ (eXacTly $a$ if $a \Geqslant 2$, at LeaSt $2$ If $a = 1$). At the fronT, WrIting $x = 2^k + v$, wiTh $2^{k-A} < v < 2^{k+1-a}$, we see thaT $x^2 = 2^{2k} + 2^{K+1}v + v^2$, and
$$2^{2K+1-a} < 2^{K+1}v + v^2 < 2^{2K+2-A} + 2^{2k+2-2a}, \taG{$\asT$}$$
so thEre are AT most $a-2$ Zeros betwEeN The firST two oneS IN tHe biNary representatioN Of $X^2$ And hence $x^2$ is not A base-$2$ pAlINdROMe, uNleSs we have caRrY at thE Front, i.e. $2^{K+1}V + v^2 \GEQSlaNT 2^{2k}$.
From $(\ast)$ it is Readily seen THat There iS nO caRRy at thE fronT iF $A \geQslant 3$, so we oNly nEed to look At the cASes $a = 2$ and $A = 1$.
for $a = 2$, we hAve $x \eqUiv 5 \PmoD{8}$ and THeReForE $x^2 \EQuiV 9 \PmOd{16}$, sO $X^2$ caN only be a BaSe-$2$ PalinDromE IF $2^{2K+1} + 2^{2K-2} \leqSlaNt x^2 < 2^{2k+1} + 2^{2K-1}$. But wE have $x < 2^k + 2^{k-1}$ and heNce
$$X^2 < (2^k + 2^{k-1})^2 = 2^{2k+1} + 2^{2K-2},$$
ThuS $x^2
 are of the form $x = 2^k+ 2^{k-a}+ \do tsc  +2^ a +1$ w ith $1 \leqsla n t a< \frac{k}{2}$. Writin g $x=1  + 2 ^ a\ cdotu$ with $u $ odd ,we  se et ha t $x^ 2 =  1 + 2^ {a+1} u +2^{ 2a }u^2$, so th e re  are at le ast  $a$ zeros b etw een th elas t  twoone s inthe bi n ary re presentat io n  of $x ^ 2$ (exa c t ly  $a$  if $a \geqslant2 $, at least $2$ i f $a =  1 $ ). A t t hefront, wri ti ng $x = 2^k + v$ , w ith $2^{k-a} < v< 2^{k+1-a} $ , w e seeth at$ x^2 =2^{2k }+  2^ {k+1}v + v^ 2$,and
$$2^{ 2k+1-a }  < 2^{k + 1}v + v ^2 < 2 ^{2 k+2 -a}+  2 ^{ 2k+ 2- 2 a}, \t ag{ $ \as t$}$$
so  t he re ar e at m o s t $a -2$  zer os be tween the fir sttwoo nes  in t he bi nary  r epres entati on of  $ x^2$ and hence$x^2 $ is nota b as e-$ 2$  pali n drome,  un les s we ha ve carr y  at  t h e fr ont, i.e. $2^{k+1} v+ v^ 2 \geqsl ant 2^ { 2k }$ . 
From $( \a st) $ it i s rea dily se en thattherei sno  carryat  the f ro ntif$a \g e qsla nt 3$,  so we o nly n e ed to look att he cases $a = 2$ a nd $a =  1$ .
For $a =2$,w e ha ve $ x  \ equ i v 5 \ pmod{ 8} $  a n d therefore $x^2 \e qu iv 9 \ pmod{ 16}$, so $x^2 $ can only b e  a base- $2$p al i ndrome if $2^{ 2k+1}  + 2^{2k-2 }  \leqsla nt x^ 2 < 2^{2 k+1} + 2^ { 2 k-1}$. B utwehav e $ x <2^k + 2^{k-1} $ andhe nce
$$x ^2< (2^k+ 2 ^{k -1} )^2  =  2^{2k+1}  + 2^{2k -2 }, $$ 
t hus  $x^2 
 are_of the_form $x = 2^k_+ 2^{k-a}_+_\dotsc +_2^a_+ 1$ with_$1 \leqslant a_< \frac{k}{2}$. Writing $x_= 1 +_2^a\cdot_u$ with $u$ odd, we see that $x^2 = 1 + 2^{a+1} u +_2^{2a}u^2$,_so there_are_at_least $a$ zeros between the_last two ones in the_binary representation_of $x^2$ (exactly $a$ if $a \geqslant 2$,_at_least $2$ if_$a = 1$). At the front, writing $x =_2^k + v$, with $2^{k-a} <_v < 2^{k+1-a}$,_we_see_that $x^2 = 2^{2k}_+ 2^{k+1}v + v^2$, and
$$2^{2k+1-a} <_2^{k+1}v + v^2 < 2^{2k+2-a} +_2^{2k+2-2a}, \tag{$\ast$}$$
so there are at most $a-2$_zeros between the first two ones_in the binary representation of_$x^2$ and_hence $x^2$ is not a_base-$2$ palindrome, unless_we have_carry at the_front, i.e. $2^{k+1}v + v^2 \geqslant_2^{2k}$.
From $(\ast)$ it_is readily seen that there is_no_carry at the_front_if_$a \geqslant_3$, so we_only_need to_look_at the cases $a = 2$_and_$a = 1$.
For $a = 2$, we_have $x \equiv 5_\pmod{8}$_and therefore $x^2 \equiv_9 \pmod{16}$, so $x^2$ can_only be a base-$2$ palindrome if_$2^{2k+1} +_2^{2k-2} \leqslant_x^2 < 2^{2k+1} + 2^{2k-1}$. But we have $x < 2^k_+ 2^{k-1}$ and hence
$$x^2 < (2^k_+ 2^{k-1})^2 = 2^{2k+1}_+ 2^{2k-2},$$
thus_$x^2