Find all five digit number $\overline{abcde}$ such that $\overline{abcde} = \overline{(ace})^2$ Find all five digit number $\overline{abcde}$ such that $$\overline{abcde} = \overline{(ace})^2$$
This question popped in my mind while solving other elementary numbers and I have been trying to solve it ever since but without any luck.
My Take : Since the digit place of $e^2$ should be equal to digit's place of $e$ , So the only possible values of $e$ are $0,1,5$ and $6$ 
Also since the First digit of the numbers are equal , We can conclude that the the only possible values of $a=1$.
Hence our number can take the following form : 
$$(1bcd0),(1bcd1),(1bcd5),(1bcd6)$$
But how do we further solve this?
Also Another interesting part of this question would be to solve for $\overline{abcd}$ such that $$\overline{abcd} = \overline{(bd)}^2$$
 A: 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):
$100^2=10000$
$1xx^2=1xxxx$
$110^2=12100$
