4. The denominator of an irreducible fraction is less than the square of the numerator by one. If 2 is added to both the numerator and the denominator, the value of the fraction will be greater than $\frac{1}{3}$. If 3 is subtracted from both the numerator and the denominator, the fraction will be less than $\frac{1}{10}$. Find this fraction, given that the numerator and denominator are single-digit numbers. Write the answer as a two-digit number, where the first digit is the numerator and the second digit is the denominator.
4. Let $\frac{m}{n}$ be the desired irreducible fraction, where $m, n$ are single-digit numbers. According to the problem, we have $n=m^{2}-1, \frac{m+2}{n+2}>\frac{1}{3} \Rightarrow 3 m+6>m^{2}+1, m^{2}-3 m-5<0$, then $\frac{3-\sqrt{29}}{2}<m<\frac{3+\sqrt{29}}{2} \Rightarrow$ $\Rightarrow 0<m<\frac{3+5.4}{2} \Rightarrow 0<m<4.2$. Therefore, $m \in\{1,2,3,4\}$. Then $m=1$ does not work because $n=0$; similarly, $m=4$ does not work because $n=15$ is a two-digit number. We check $m=2$, then $n=3$ and the fraction will be $\frac{2}{3}$, which is impossible according to the problem. The only option left is $m=3$, then $n=8$, The fraction $\frac{3}{8}$ satisfies all the conditions of the problem. Answer: $\{38\}$.