Finding number $\overline{ATOM}$ such that $\sqrt{\overline{ATOM}}=A+\overline{TO}+M$ $\overline{ATOM}$ is four digits number such that
$\sqrt{\overline{ATOM}}=A+\overline{TO}+M$. How to find the numbers  ?
I am sure that $M\neq2,3,7,8$ because it is a perfect square.
I wrote $\overline{ATOM}=M+10O+100T+1000T$ and r.h.s as $M+O+10T+A$ I tried some algebraic calculations, but I  can't go further more. I have no clue to solve this problem. 
If you can give an idea to solve, I am thankful. 
 A: Let $n=\overline{A}+\overline{TO}+\overline{M}, n^2=\overline{ATOM}$.  Clearly these must have the same digital root so $n^2\equiv n \bmod 9$.  Since $9$ has only one distinct prime factor $n\equiv 0$ or $n\equiv 1 \bmod 9$.
The tens digit of $n$ must be greater than $2$ for $n^2$ to have four digits without an initial zero. We cannot have $n$ ending with $0$, else $\overline{OM}=00$ forcing $A=0$. 
Nor does $n$ end with $5$ since that leads to $\overline{OM}=25$ and $A=8$, but there are no squares having the form $8T25$.
So we try $n=36, 54, 63, 72, 81, 99$ among multiples of $9$; and $n=37, 46, 64, 73, 82, 91$ among numbers one greater than a multiple of $9$.  $n=36$ and $n=82$ work.
A: Not a very efficient solution: Let $x=\overline{TO}$. Then $$1000A+10x+M=x^2+2x(A+M)+(A+M)^2$$ and $$x^2+(2A+2M-10)x+(A+M)^2-(1000A+M)=0$$Discriminant must be a perfect square, that is $$\Delta_x^{'}=9(110A-M)+25=k^2$$ Thus $k=9l+4$ or $k=9l+5$. 
They lead to $110A-M+1=9l(9l+8)$ and $110A-M=9l(9l+10)$, respectively. Consider modulo $9$ and find possible values of $A$ from possible values of $M$ as you and @symplectomorphic ruled out them. Then find $x$. You have 10 cases to check. 
