Please verify my proof of: There is no integer $\geq2$ sum of squares of whose digits equal the integer itself. While going to sleep, I just started thinking about numbers, their squares, cubes and after thinking for about $20$ minutes I got that:

There is no integer (having any numbers of digits) except $0$ and $1$, sum of squares of whose digits is equal to number itself.

Isn't it interesting ??
Now I came out blanket and started writing everything on paper (Or started finding a proof of what I have got).
MY WORK
Suppose there exist an integer $a_0a_1a_2.......a_{n-1}$  having n digits.
Firstly, we can easily exclude negative integers from the race as the number will be negative and sum of squares will be positive (we can never equate negative and positive).
Now comes the turn of non negative integers.
As we have assume that there exist an integer which satisfy our condition so, it should yield us:
$$10^{n-1}a_0+10^{n-2}a_1+.........10a_{n-2}+a_{n-1}={a_0}^2+{a_1}^2+.......{a_{n-2}}^2+{a_{n-1}}^2$$
Which on further solving becomes,
$$a_0(10^{n-1}-a_0)+a_1(10^{n-2}-a_1)+.........+a_{n-2}(10-a_{n-2})+a_{n-1}(1-a_{n-1})=0$$
$$a_0(10^{n-1}-a_0)+a_1(10^{n-2}-a_1)+.........+a_{n-2}(10-a_{n-2})=a_{n-1}(a_{n-1}-1)$$
Now, the hardest part for me comes:
The terms of left are all positive and they increase as we keep on going towards left.
We can only compare the terms $$a_{n-1}(a_{n-1}-1)$$ and $$a_{n-2}(10-a_{n-2})$$ because the shortest of other terms is too far from comparison.
Now if we make some comparison, we will have $$a_{n-1}(a_{n-1}-1)=a_{n-2}(10-a_{n-2})$$
Now let's form a table for the function on RHS:
$$\begin{array}{c|c}
a_{n-2}&            a_{n-2}(10-a_{n-2})\\\hline
0&                         0\\
1&                         9\\
2&                         16\\
3&                         21\\
4&                         24\\                         
5&                         25\\
6&                         24\\
7&                         21\\
8&                         16\\
9&                         9
\end{array}$$
Again, we form a table and this time for the function on LHS:
$$\begin{array}{c|c}
a_{n-1}&                       a_{n-1}(a_{n-1}-1)\\\hline
0&                                  0\\
1&                                  0\\
2&                                  2\\
3&                                  6\\
4&                                 12\\
5&                                 20\\
6&                                 30\\
7&                                 42\\
8&                                 56\\
9&                                 72
\end{array}$$
Now we have all possible values of  $a_{n-1}(a_{n-1}-1)$ and  $a_{n-2}(10-a_{n-2})$ and A quick look at tables yield that the only common value $a_{n-1}(a_{n-1}-1)$ and  $a_{n-2}(10-a_{n-2})$ have is $0$. 
So the condition is followed when ($a_{n-1}$ ,$a_{n-2}$ )=($0,0$) and ($1,0$). Or $1$ and $0$ are the only numbers which follow the condition.{ There is no integer (having any numbers of digits) except $0$ and $1$, sum of squares of whose digits is equal to number itself is proved}.
I hope you guys will understand that such questions are hard to write (specially for someone like me who is a beginer at Mathjax). So, if you have any problem in understanding, leave comment. I shall be thankful if someone can verify my proof or can give a new one (with completely different approach). You can suggest modification in my work or give suggestion that where it can be improved.Thanks
 A: Your argument is good, though a little vague at the end.  I would tighten it up by remarking that $$a_{n-k}>0\implies a_{n-k}\left(10^{n-1-k}-a_{n-k}\right)>1\times \left(10^{n-1-k}-9\right)$$ which is greater than $99$ for all the terms you wish to discard. In that way, you reduce the problem to two digit cases, and you resolve those correctly.
An algebraically simpler argument:  If your integer has $n$ digits, then the sum of squares can be at most $81n$.  But we have $$10^{n-1}≤81n\implies n≤3$$ so your number can have at most $3$ digits.  Noting that $81\times 3= 243$ we see that in fact we only have to check up up $243$. This can easily be checked by (mechanically assisted) hand.
A: The problem requires
$$
\begin{align}
0
&=\sum_{k=0}^n\left(10^kd_k-d_k^2\right)\\
&=\sum_{k=0}^nd_k\left(10^k-d_k\right)
\end{align}
$$
The only negative term can be $d_0(1-d_0)\ge-72$.
For $k\ge2$, if $d_k\ne0$, then $d_k\left(10^k-d_k\right)\ge10^k-1\ge99$.
Therefore, only $d_0,d_1$ can be non-zero. So we just need to find $d_0,d_1$ so that
$$
d_1(10-d_1)=d_0(d_0-1)
$$
However
$$
d_1(10-d_1)\in\{0,9,16,21,24,25\}
$$
while
$$
d_0(d_0-1)\in\{0,2,6,12,20,30,42,56,72\}
$$
The only way they can be equal is if both are $0$ and that requires $d_1=0$ and $d_0\in\{0,1\}$.
That is, the only solutions are $0$ and $1$.
Your proof was a bit difficult to read, but it looks similar.
