We can minimize trial and error with some clever use of modular arithmetic.
Let $N=100a+10c+e$ be the square root. Thus $N^2\equiv e^2$ and we require also $N^2\equiv e\bmod 10$. Therefore $e^2\equiv e$ forcing $e\in\{0,1,5,6\}$.
We also know that $(100a)^2<10000(a+1)$ or $a^2<a+1$ forcing $a=1$. Then $N^2<20000$ but $145^2>140×150=21000$, therefore $N<145$. This result together with the earlier constraint on $e$ leaves only eighteen candidates, which can be exhaustively searched with little trouble; but we can do even better than that.
Consider the case $e=0$. Then $N=100+10c$ (with $a=1$) and $N^2=10000+2000c+100c^2$. For the hundreds digit in $N^2$ to match $c$ we must then have $c^2\equiv c\bmod 10$. This constraint admits$c\in\{0,1,5,6\}$, but only $0$ and $1$ satisfy the bounty $N<145$ which implies $c\le 4$. Thereby we identify
$100^2=10000$
$110^2=12100$
For $e=1$ we have
$N^2=10000+2000c+100(c^2+2)+20c+1$
With $c\le 4$, $20c+1<100$ and thus the hundreds digit is $\equiv c^2+2\bmod 10$. Therefore we must satisfy
$c^2-c+2\equiv 0\bmod 10$
which has a discriminant that is not a quadratic residue $\bmod 5$. So nobody's home here.
The cases $e=5$ and $e=6$ are left to the reader; they are handled similarly to $e=1$ as described above. For these cases $N<145$ implies $c\le 3$ which will then fix the hundreds digit of $N^2$ as $\equiv c^2+c$ ($e=5$) or $\equiv c^2+c+2$ ($e=6$). We will then get only one additional solution which the reader can find. I list the complete solution set as (with $x$ digits to be filled in):
$100^2=10000$
$1xx^2=1xxxx$
$110^2=12100$
Let us consider the permutation
$$\sigma ={\begin{pmatrix}1&2&3&4&5&6&7\\2&3&4&5&6&7&1\end{pmatrix}}$$
That is, $\sigma(i)=i+1$ for all $i$, $1\le i\le 6$ and $\sigma(7)=1$.
Note that $\sigma(i)\equiv_7i+1$ for all $i$.
Call a number that can be obtained by permuting the digits of $1234567$ a good number.
For a good number $n$ that is $d_6d_5d_4d_3d_2d_1d_0$ in decimal notation, let $\sigma(n)$ be $\sigma(d_6)\sigma(d_5)\sigma(d_4)\sigma(d_3)\sigma(d_2)\sigma(d_1)\sigma(d_0)$ in decimal notation. So $\sigma$ maps a good number to a good number.
Since $\sigma^7$ is the identity map, all good numbers are grouped into disjoints cosets, each of which is $\{n, \sigma(n), \sigma^2(n), \sigma^3(n), \sigma^4(n), \sigma^5(n), \sigma^6(n)\}$ for some good number $n$. (For example, $\{1354762, 2465173, 3576214, 4617325, 5721436, 6132547, 7243651\}$.)
$$\sigma(n)-n\equiv_71111111=7*11*13 * 1110 + 1\equiv_71$$
So, there is exactly one number in each coset that is divisible by 7.
Since there are $\frac{7!}7=720$ cosets, there are $720$ good numbers that are divisible by $7$.
Best Answer
We convert the given equation to the following form. $$1000(AB)+(CDE)=4(AB)^2+(CDE)^2. $$
Now consider the following quadratic equation. $$(CDE)^2-(CDE)+4(AB)^2-1000(AB)=0$$
The two roots of this equation are, $$(CDE)=\dfrac{1\pm \sqrt{1-16(AB)^2+4000(AB)}}{2}.$$
Since $(AB)$ is a 2-digit number, $10\le (AB)\le 99$. For all these values of (AB), $$\sqrt{1-16(AB)^2+4000(AB)} \gt 1$$.
Therefore, we have, $$(CDE)=\dfrac{1+ \sqrt{1-16(AB)^2+4000(AB)}}{2}.$$
Now, we have to find out the values of $(AB)$, for which the expression under the square-root sign is a square. They turn out to be, $$(AB)=19\qquad\text{and}\qquad (AB)=70.$$*
The corresponding values of $(CDE)$ are, $$(CDE)=133\qquad\text{and}\qquad (CDE)=225.$$
*Here is how to get these values of $\overline{AB}$:
First complete the square in the discriminant. This gives
$1-16(\overline{AB})^2+4000\overline{AB}=250001-(4\overline{AB}-500)^2.$
Thus we must have
$250001=m^2+(4\overline{AB}-500)^2$.
Since $250001=500^2+1^2$ where the squares are relatively prime, the odd number must be a product of $4n+1$ primes only. Trial divisions then yield the factorization
$250001=53^2×89=(7^2+2^2)^2(8^2+5^2).$
We now consider Gaussian-integer products of the form
$(7\pm2i)(7\pm2i)(8\pm5i).$
Each product has real and imaginary parts whose squares sum to $250001$. Wlog we may take the first factor to be specifically $7+2i$, and we find three combinations giving different squares:
$(7+2i)(7+2i)(8+5i)=220+449i$
$(7+2i)(7+2i)(8-5i)=500-i$
$(7+2i)(7-2i)(8\pm5i)=424\pm265i.$
The first product above gives
$250001=220^2+449^2,$
from which $4\overline{AB}-500$ which is a multiple of $4$ must be $\pm220$. We choose the negative possibility to force $\overline{AB}<100$, so we then find
$\color{blue}{\overline{AB}=70.}$
The third product similarly gives the other nonzero value for $\overline{AB}$:
$\color{blue}{\overline{AB}=19.}$