$111111, 111112, 111121, 111211, 112111, 121111, 211111, 1111111, 1111112, 1111121,$ $ 1111211, 1121111, 1211111, 2111111, 11111111, 111111111$

$111111, 111112, 111121, 111211, 112111, 121111, 211111, 1111111, 1111112, 1111121,$ $ 1111211, 1121111, 1211111, 2111111, 11111111, 111111111$

Since rightmost digit is greater than leftmost digit, it is $\ge 2$. If rightmost digit is $4$, $A^2$ ends with $6$ while $Z(A)^2$ begins by $16-24$, impossible If rightmost digit is $5$, $A^2$ ends with $5$ while $Z(A)^2$ begins by $25-35$, impossible If rightmost digit is $6$, $A^2$ ends with $6$ while $Z(A)^2$ begins by $36-48$, impossible If rightmost digit is $7$, $A^2$ ends with $9$ while $Z(A)^2$ begins by $49-63$, impossible If rightmost digit is $8$, $A^2$ ends with $4$ while $Z(A)^2$ begins by $64-80$, impossible If rightmost digit is $9$, $A^2$ ends with $1$ while $Z(A)^2$ begins by $81-99$, impossible So leftmost digit of $A$ is $\le 2$ and $A^2$ has $2n-1$ digits Q.E.D.

If rightmost digit is $3$, $A^2$ ends with $9$ and so $Z(A^2)=Z(A)^2$ begins by $9$ And so $Z(A)\in[\sqrt{9}10^{n-1},\sqrt{10}10^{n-1})=[30000...,31622...)$ So next digit from the right is $1$ (since $Z(A)<32000...$) So $A$ ends with $13$, $A^2$ ends with $69$ and $Z(A^2)=Z(A)^2$ begins by $96$ And so $Z(A)\in[\sqrt{9.6}10^{n-1},\sqrt{9.7}10^{n-1})=[3098...,31144...)$ So next digit from the right is $1$ again (since $Z(A)<31200...$) So $A$ ends with $113$, $A^2$ ends with $769$ and $Z(A^2)=Z(A)^2$ begins by $967$ And so $Z(A)\in[\sqrt{9.67}10^{n-1},\sqrt{9.68}10^{n-1})=[31096...,311126..)$ So next digit from the right is $1$ again (since $Z(A)<311200...$) So $A$ ends with $1113$, $A^2$ ends with $8769$ and $Z(A^2)=Z(A)^2$ begins by $9678$ And so $Z(A)\in[\sqrt{9.678}10^{n-1},\sqrt{9.679}10^{n-1})=[31109...,3111109..)$ And so no such good number (since sixth digit from the right would be zero) Q.E.D.

$A^2$ ends with $4$ and so $Z(A^2)=Z(A)^2$ begins by $4$ and so $Z(A)\in[\sqrt{4}10^{n-1},\sqrt{5}10^{n-1})=[20000...,2236...)$ So next digit from the right is $1$ or $2$ (since $Z(A)<23000...$) 3.1) if rightmost digits are $22$, no such good number ProofSo $A$ ends with $22$, $A^2$ ends with $84$ and $Z(A^2)=Z(A)^2$ begins by $48$ And so $Z(A)\in[\sqrt{4.8}10^{n-1},\sqrt{4.9}10^{n-1})=[21908...,22135...)$ So next digit from the right is $1$ (since $Z(A)<22200...$) So $A$ ends with $122$, $A^2$ ends with $884$ and $Z(A^2)=Z(A)^2$ begins by $488$ And so $Z(A)\in[\sqrt{4.88}10^{n-1},\sqrt{4.89}10^{n-1})=[22090...,221133...)$ So next digit from the right is $1$ again (since $Z(A)<221200...$) So $A$ ends with $1122$, $A^2$ ends with $8884$ and $Z(A^2)=Z(A)^2$ begins by $4888$ And so $Z(A)\in[\sqrt{4.888}10^{n-1},\sqrt{4.889}10^{n-1})=[221088...,2211108...)$ And so no such good number (since sixth digit from the right would be zero) Q.E.D. 3.2) if rightmost digits are $12$ So $A$ ends with $12$, $A^2$ ends with $44$ and $Z(A^2)=Z(A)^2$ begins by $44$ And so $Z(A)\in[\sqrt{4.4}10^{n-1},\sqrt{4.5}10^{n-1})=[20976...,21213...)$ So next digit from the right is $1$ or $2$ (since $Z(A)<21300...$) 3.2.1) if rightmost digits are $212$, no such good number ProofSo $A$ ends with $212$, $A^2$ ends with $944$ and $Z(A^2)=Z(A)^2$ begins by $449$ And so $Z(A)\in[\sqrt{4.49}10^{n-1},\sqrt{4.50}10^{n-1})=[21896...,21213...)$ So next digit is $1$ (since $Z(A)<21220...$) So $A$ ends with $1212$, $A^2$ ends with $8944$ and $Z(A^2)=Z(A)^2$ begins by $4498$ And so $Z(A)\in[\sqrt{4.498}10^{n-1},\sqrt{4.499}10^{n-1})=[212084...,212108...)$ And so no such good number (since fifth digit from the right would be zero) Q.E.D. 3.2.2) if rightmost digits are $112$, two solutions SolutionsSo $A$ ends with $112$, $A^2$ ends with $544$ and $Z(A^2)=Z(A)^2$ begins by $445$ And so $Z(A)\in[\sqrt{4.45}10^{n-1},\sqrt{4.46}10^{n-1})=[21095...,211187...)$ So next digit from the right is $1$ (since $Z(A)<21120...$) So $A$ ends with $1112$, $A^2$ ends with $6544$ and $Z(A^2)=Z(A)^2$ begins by $4456$ And so $Z(A)\in[\sqrt{4.456}10^{n-1},\sqrt{4.457}10^{n-1})=[211092...,2111160...)$ So next digit from the right is again $1$ (since $Z(A)<211120...$) So $A$ ends with $11112$, $A^2$ ends with $76544$ and $Z(A^2)=Z(A)^2$ begins by $44567$ And so $Z(A)\in[\sqrt{4.4567}10^{n-1},\sqrt{4.4568}10^{n-1})=[2111089...,21111134...)$ So next digit from the right is again $1$ (since $Z(A)<211112...$) So $A$ ends with $111112$ and the only six digit in this category could be $111112$ which indeed fits If $A$ has more than 6 digits, then $A^2$ ends with $876544$ and $Z(A^2)=Z(A)^2$ begins by $445678$ And so $Z(A)\in[\sqrt{4.45678}10^{n-1},\sqrt{4.45679}10^{n-1})=[21111087...,2111111089...)$ So next digit from the right is $1$ again since $Z(A)<2111112...$) and no possible eighth digit (which would be zero) And indeed $A=1111112$ fits