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 :)