43. Find a four-digit number that is a perfect square, where the digit in the thousands place is the same as the digit in the tens place, and the digit in the hundreds place is 1 more than the digit in the units place.
43. The desired number is $1000a + 100(b+1) + 10a + b = 1010a + 101b + 100$. $1010a + 101b + 100 = x^2$ or $101(10a + b) = (x-10)(x+10)$.

The left side has a factor of 101, so the right side must also have such a factor. Since $x$ is a two-digit number, $x-10 < 101$. Therefore, $x+10 = 101$, from which $x = 91$.

The desired number is 8281.