i think this is from ISL the proof isnt elegant at all. if i remember correctly, it just involves considering lots and lots of cases, relating to the last digit of a. We get contradictions for most cases but i think we get 3 infinite solutions at the end.

I remember one infinite solution and some (three?) finite ones, but I am not sure. I don't know the solution any more (I have completely messed up the problem during the exam, and later, when Gronau showed us the solution, I have been sleeping). Darij

In what follows, we assume $a$ has $n+1$ digits, meaning $10^n \le a < 10^{n+1}$. In particular, $10^{2n} \le d < 10^{2n+2}$. First, we analyze the last digit of $a$, say $x$. Suppose \begin{align*} a &= \overline{\dots x} \\ b &= \overline{x \dots} \\ c &= b^2 = \overline{y \dots} \\ d &= \overline{\dots y} \end{align*}so that $y = x^2 \pmod{10}$ (here $\dots$ denotes some irrelevant digits). Claim: $x \neq 0$. Proof. Otherwise, $b < 10^n$, $c < 10^{2n}$, so $d < 10^{2n}$. $\blacksquare$ Next, we make a large table by noting that if we know the leading digit of $b$, we can find all possible leading digits of $c = b^2$, hence all possible final digits of $d$. \[ \begin{array}{r|ccccc|cccc} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline x^2 \pmod{10} & 1 & 4 & 9 & 6 & 5 & 6 & 9 & 4 & 1 \\ \text{possible first digits of } \overline{x\dots}^2 & 123 & 45678 & 91 & 12 & 23 & 34 & 456 & 678 & 89 \end{array} \]From this, we conclude that $x \in \{1,2,3\}$, and we must have $y = x^2$. In particular, $10^{2n} \le c < 10^{2n+1}$, hence $10^{2n} \le d < 10^{2n+1}$. Now we proceed to do the calculation: set \begin{align*} a &= 10m + x \\ b &= 10^n x + m \\ c &= 10^{2n} x^2 + 2 \cdot 10^n \cdot mx + m^2 \\ d &= 10\left( 2 \cdot 10^n \cdot mx + m^2 \right) + x^2 \end{align*}Since $a^2 = d$, we have \begin{align*} 100 m^2 + 20 mx + x^2 &= 10\left( 2 \cdot 10^n \cdot mx + m^2 \right) + x^2 \\ 10m + 2x &= 2 \cdot 10^n \cdot x + m \\ m &= 2x \cdot \frac{10^n-1}{9} = 2x \cdot \underbrace{1\dots1}_n \end{align*}In other words, $a$ should be one of the following shapes: $a = 2\dots21$, $b = 4\dots42$, $c = 6\dots63$. A quick check now shows the answers are: \[ a = \underbrace{2\dots2}_{n \ge 0}1, \qquad a = 2, \qquad a = 3. \]

R. Let $a$ be a number that works. Let $a=10a_0+a_1$ where $a_0$ is a nonnegative integer and $b$ is a nonnegative integer from $0$ to $9$ inclusive. Then $b$ will have first digit $a_1.$ $d$ must have last digit equivalent to $a_1^2$ mod $10.$ So its first digit is equivalent to $a_1^2$ mod $10,$ all assuming $a_0\ge 1.$ We'll deal with $a_0=0$ later. Now, we can compare the possible values of $a_1.$ For $a_1=4,5,6,\dots,9$ the possible values of the first digit of $c$ does not compute based on $b$ does not compute with the possible values of the first digit of $c$ based on $d.$ Thus, $a_1=1,2,3.$ Suppose $a$ has $n$ digits. Then $10^{n-1}\le a\le 10^{n}-1.$ We have $b=10^n\cdot a_1+a_0.$ Thus, \[c=10^{2n}\cdot a_1^2+a_0^2+2\cdot 10^n\cdot a_0a_1\]We have $d=100a_0^2+20a_0a_1+a_1^2=10(10a_0^2+2a_0a_1)+a_1^2.$ Thus, \[c=10^{2n}a_1^2+10a_0^2+2a_0a_1\]Thus, we have $9a_0=2(10^n-1)a_1$ so $a_0=2(111111...111)a_1$ with $n$ ones. Thus, $a$ is in the form $22222...21,$ $4444....42$ or $6666...63.$ One easily checks that the working ones are $2222...221,2,3$. Now for $a_0=0:$ This is easy to prove impossible by the number of digits.

what the bad Write $A=10X+Y$, where $Y$ is the units digit, and $X$ has $d$ digits. Then: $A = 10X+Y$ $B = 10^dY+X$ $C=10^{2d}Y^2+10^d2XY+X^2$ $A^2=10^2X^2+10\cdot 2XY+Y^2$ To solve the problem, it is first necessary to narrow the possibilities for $Y$ with bounding. $Y=0$: Doesn't work. Then $D$ has units digit $0$, so that $C$ has first digit $0$, impossible. $Y=1$: Works. Then $D$ has units digit $1$, so that $C$ has first digit $1$. Since $B$ has first digit $1$, this is potentially achieveable. $Y=2$. Works. Then $D$ has units digit $4$, so that $C$ has first digit $4$. Since $B$ has first digit $2$, this is potentially achieveable. $Y=3$. Works. Then $D$ has units digit $9$, so that $C$ has first digit $9$. Since $B$ has first digit $3$, this is potentially achieveable. $Y=4$. Doesn't work. $D$ has units digit $6$, so $C$ has first digit $6$. But $B$ has first digit $4$, which would imply that the first digit of $C$ is either $1$ or $2$, a fail. $Y=5$. Doesn't work. $D$ has units digit $5$, so $C$ has first digit $5$. But $B$ has first digit $5$, which would imply that the first digit of $C$ is either $2$ or $3$, a fail. $Y=6$. Doesn't work. $D$ has units digit $6$, so $C$ has first digit $6$. But $B$ has first digit $6$, which would imply that the first digit of $C$ is either $3$ or $4$, a fail. $Y=7$. Doesn't work. $D$ has units digit $9$, so $C$ has first digit $9$. But $B$ has first digit $7$, which would imply that the first digit of $C$ is either $4$, $5$, or $6$, a fail. $Y=8$. Doesn't work. $D$ has units digit $4$, so $C$ has first digit $4$. But $B$ has first digit $8$, which would imply that the first digit of $C$ is either $6$, $7$, or $8$, a fail. $Y=9$. Doesn't work. $D$ has units digit $1$, so $C$ has first digit $1$. But $B$ has first digit $9$, which would imply that the first digit of $C$ is either $8$ or $9$, a fail. With this check, we can prove that the $10^{2d}Y^2$ term in the sum creating $C$ is precisely the largest digit. Notice that \[10^d\le A,B<10^{d+1}\implies 10^{2d}\le C<10^{2d+2}.\]And yet, for $Y=1$ and $Y=2$, we find $B<3\cdot 10^d$, hence $C<10^{2d+1}$, so that the $10^{2d}Y^2$ represents the first digit of $C$. With $Y=3$, the condition remains that $C$ has first digit $9$, so that $10^{2d}Y^2$ represents the first digit of $C$; if $C$ was at least $10^{2d+1}$ here, it would only have first digit $1$. Hence $D$ is precisely equal to: \[C=10^{2d}Y^2+10^d2XY+X^2\implies D=Y^2+10(10^d2XY+X^2)\]so that \[Y^2+10(10^d2XY+X^2)=D=A^2=10^2X^2+10\cdot 2XY+Y^2\]\[10(10^d2XY+X^2)=10^2X^2+10\cdot 2XY\]\[10^d2XY+X^2=10X^2+2XY\]\[10^d2Y+X=10X+2Y\]\[2Y(10^d-1)=9X.\] Hence we have three infinite sets of solutions: \[A=\underbrace{2\dots2}_{n \ge 0}1,\qquad A=\underbrace{4\dots4}_{n \ge 0}2,\qquad A=\underbrace{6\dots6}_{n \ge 0}3.\]For the last two cases, we may check that for $n\ge 1$ the "first digit" condition of $C$ (that the first digit is $4$ and $9$, respectively) fails. Hence only $A=2$ and $A=3$ work in these cases. In the first case, the first digit condition of $C$ holds, and the number of digits of $C$ "matches up," so the solution set is valid. The final answer is: \[A=\underbrace{2\dots2}_{n \ge 0}1,\qquad A=2,\qquad A=3.\]We are done. $\blacksquare$