prove that there are no number such that $\overline{ab\ldots}=a^2+b^2+\ldots$ Prove that a natural number with at least 2 digits cannot be written like a sum with the the power of digits equal $2$. 
What I want to say: $$\overline{ab}\neq a^2+b^2.$$
What I have done: 
$$10a+b=a^2+b^2 $$ or 
$$a(a-10)=b(1-b).$$
$b(1-b)=2k$ so $a(a-10)=2k$ and this is possible only when $a=2q.$ 
so: $2 \cdot 8 =b(b-1)$ or $4\cdot 6=b(b-1)$ and this is not possible. 
final conclusion for the number $\overline{ab}$ is ok, but what can I do for number formatted with $3,4, \ldots$ digits ? 
thanks :) 
 A: Assume that
$$\overline{a_n...a_0}=a_n^2+..+a_0^2 \,.$$
Then
$$10^n \leq \overline{a_n...a_0} = a_n^2+..+a_0^2  \leq 9^2+...+9^2 =81(n+1)$$
It is easy to prove (by induction or using calculus) that for $n \geq 3$ we have
$$10^n >81(n+1) \,.$$
This implies that $n \leq 2$, which completes your proof.
Edit When $n=2$ we need to solve
$$100a+10b+c=a^2+b^2+c^2$$
This implies 
$$a(100-a)=b^2-10b+c^2-c$$
If $a \neq 0$, the left hand side is at least $91a \geq 91$, while the RHS is at most $c(c-1) \leq 72$, so they cannot be equal.
A: For more than 3 digits, if there are $n$ digits the sum of squares is at most $81n$ while the number is at least $10^{n-1}$ and you are done.  For specifically $3$ digits, the maximum sum of squares would be $3 \cdot 81=243$, but then the leading digit is no more than $2$, so the maximum sum is no more than $4+162=166$, so the leading digit is $1$, so the maximum sum is $163$, so the maximum sum of squares is $1+36+81=118$, so the maximum sum of squares is $1+1+81=83$, failure.
A: For an $n$ digit number, the left hand side is $\ge 10^{n-1}$ and the right hand side is $\le 81n$.  Therefore, only $n\le 3$ has to be considered.
This is only a very finite problem - if one is too lazy to find a clever argument, one can simply check the less than 1000 candidate numbers by computer.
