45. (9th grade) Find a four-digit number that is a perfect square, knowing that the first two digits, as well as the last two, are equal to each other.
45. The desired number can be written as: $N=\overline{a a b} b$ or $N=1100 a+11 b=$ $=11(99 a+a+b)$. For the desired number to be a perfect square, it is necessary that the factor 11 appears in the number $N$ twice, i.e., the sum $a+b$ must be divisible by $11.1<a+b<18$, hence, $a+b=11$. Then $N=11^{2}(9 a+1)$. The binomial $9 a+1$ must be a square of some number. This is possible only when $a=7$. Then $b=4, N=7744=88^{2}$.