151. Find a four-digit number that is equal to the square of the sum of two two-digit numbers formed by the first two and the last two digits of the number.
151. Let the two-digit number formed by the first two digits of the desired number $N$ be denoted by $a$, and the number formed by its last two digits be denoted by $b$. In this case, $N=100a + b$, so the condition of the problem gives

$$
100a + b = (a + b)^2
$$

or

$$
99a = (a + b)^2 - (a + b) = (a + b)(a + b - 1)
$$

Thus, the product $(a + b)(a + b - 1)$ must be divisible by 99. Let's now consider the individual possible cases.

$1^{\circ} . a + b = 99k, a + b - 1 = \frac{a}{k}$. Since $a$ and $b$ are two-digit numbers, $k \leqslant 2$, and the case $k=2$ immediately leads to values of $a=99, b=99$ that do not satisfy the main equation (*). Therefore, we must assume that

$$
\begin{aligned}
& k=1, a + b = 99, a = a + b - 1 = 98, N = 9801 = (98 + 1)^2. \\
& 2^{\circ} \cdot a + b = 11m, a + b - 1 = 9n, mn = a. In this case, we have
\end{aligned}
$$

$m$ gives a remainder of 5 when divided by 9 (a direct check shows that if $m$ gave any other remainder when divided by 9, then $11m - 1$ would not be divisible by 9). Thus, $m = 9t + 5$, from which it follows that $9n = 99t + 54, n = 11t + 6$. Now we have:

$$
a = mn = (9t + 5)(11t + 6) = 99t^2 + 109t + 30
$$

Since $a$ is a two-digit number, it immediately follows that $t = 0$. Therefore, $a = 30, a + b = 11m = 55, b = 25, N = 3025 = (30 + 25)^2$.

$3^{\circ} . a + b = 9m, a + b - 1 = 11n, mn = a$. An investigation similar to the one we conducted in case $2^{\circ}$ gives the unique answer $N = 2025 = (20 + 25)^2$.

$4^{\circ} . a + b = 33m, a + b - 1 = 3n$ or $a + b = 3m, a + b - 1 = 33n$, which is impossible, as $a + b$ and $a + b - 1$ are coprime.

$$
5^{\circ} . a + b - 1 = 99k, a + b = \frac{a}{k}. \quad \text{In this case, we have}
$$

$a + b - 1 = 99, a + b = 100, a = \frac{(a + b)(a + b - 1)}{99} = 100$, which is impossible. Therefore, the conditions of the problem are satisfied only by the numbers 9801, 3025, and 2025.