3. Three-digit number $\overline{a b c}=a^{2}+1+(\overline{b c})^{2}$. Then $\overline{a b c}=$
3.726 .

From $100 a+\overline{b c}=a^{2}+(\overline{b c})^{2}+1$, we have $(\overline{b c}-1) \overline{b c}=100 a-a^{2}-1$.
Obviously, the left side is even,
so the right side $a$ must be odd to make both sides equal.
When $a=1$, $(\overline{b c}-1) \overline{b c}=98=2 \times 7^{2}$;
When $a=3$, $(\overline{b c}-1) \overline{b c}=290=2 \times 5 \times 29$;
When $a=5$, $(\overline{b c}-1) \overline{b c}=474=2 \times 3 \times 79$;
When $a=7$, $(\overline{b c}-1) \overline{b c}=650=26 \times(26-1)$;
When $a=9$, $(\overline{b c}-1) \overline{b c}=818=2 \times 409$.
Only when $a=7$, $\overline{b c}=26$ satisfies.
Therefore, $\overline{a b c}=726$.