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 :) 

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.
