* Given a four-digit number $N$ is a perfect square, and each digit of $N$ is less than 7. Increasing each digit by 3 results in another four-digit number that is also a perfect square. Find $N$.
Let $N=a \cdot 10^{3}+b \cdot 10^{2}+c \cdot 10+d$, where integers $a, b, c, d$ satisfy $1 \leqslant a \leqslant 6$, $0 \leqslant b, c, d \leqslant 6$. From the problem, we know $N=n^{2} \leqslant 6666$, so $n \leqslant 81$. Also, $n^{2}+3333=m^{2}$, which means $(m+n)(m-n)=3 \times 11 \times 101$. Since $m+n>m-n, n \leqslant 81$, we have $m+n=101$, $m-n=33$, thus $n=34, N=1156$.