Problem 1

When $a+b\ge10$, adding $10(a+b)$ to $100a+b$ causes a carry: the hundreds digit becomes $a+1$ and the tens digit becomes $a+b-10$ (with $b$ unchanged as the units digit). The digit-square condition now reads $(a+1)^2+(a+b-10)^2+b^2=10a+b$, which expands to $2a^2+2ab+2b^2-28a-21b+101=0$. The same kind of parity argument as before (writing $b=2B+1$ and checking evenness) shows $b$ must be odd, and in fact $b=3$ or $b=7$.