2. Determine all four-digit natural numbers $\overline{a b c d}$, where different letters correspond to different digits, for which the following holds:

$$
\overline{a b c d}=d^{a^{2}}+d^{b+c}-d^{a+b+c}-d^{a}+1
$$
2. It is obvious that $d>1$. Since the number on the right side is an odd number, based on the left side, it follows that $d$ is an odd digit. Additionally, $d \neq 5$, because otherwise the left side would be divisible by 5, but the right side would not. Therefore, $d \in \{3,7,9\}$. Furthermore, we have $a>1$, because otherwise the right side would be negative.

Let's prove that $a^{2} \geqslant a+b+c$. Assume the opposite: $a^{2} \leqslant a+b+c-1$. Then we have

$$
\begin{aligned}
d^{a^{2}} + d^{b+c} - d^{a+b+c} - d^{a} + 1 & (d-1) d^{a^{2}-a-1} + 1 > d^{a^{2}-a-1}
\end{aligned}
$$

If $a^{2} > a+b+c$, it follows that $a+b+c \leqslant a^{2}-1$, so we get:

$$
d^{a^{2}} + d^{b+c} - d^{a+b+c} - d^{a} + 1 > d^{a^{2}} - d^{a^{2}-1} - d^{a} + 1 > (d-2) d^{a^{2}-1} + 1 > d^{a^{2}-a-1}
$$

Thus, we have certainly bounded the right side of the equation from below by $d^{a^{2}-a-1}$. However, for $a \geqslant 4$, we have $d^{a^{2}-a-1} \geqslant 3^{11} > 9999$, and for $a=3$, it follows that $d \in \{7,9\}$ (since $a$ and $d$ are different digits), so we have $d^{a^{2}-a-1} \geqslant 7^{6} > 9999$. Altogether, for $a \geqslant 3$, the number on the right side cannot be a four-digit number, so only $a=2$ remains.

For $a=2$, it follows that $2 \geqslant b+c$, and since $b$ and $c$ are different digits, it must be $\{b, c\} = \{0,1\}$. Then $3 \mid a+b+c$, so $3 \nmid d$ because otherwise the number on the left side would be divisible by 3 (since the sum of its digits is divisible by 3) but the number on the right side would not. Therefore, the only remaining possibility is $d=7$, and we calculate:

$$
\overline{a b c d} = 7^{2^{2}} + 7^{1} - 7^{3} - 7^{2} + 1 = 2401 + 7 - 343 - 49 + 1 = 2017
$$

which is the only solution to the problem.

![](https://cdn.mathpix.com/cropped/2024_06_04_f254ba4f25ce7f8d9df1g-37.jpg?height=257&width=733&top_left_y=1597&top_left_x=191)

Ok 2017 2A 3