Determine all triples $(a, b, c)$ of natural numbers $a, b$, and $c$ for which

$$
a^{b+20}(c-1)=c^{b+21}-1
$$

is satisfied.

(Walther Janous)

Answer. $\{(1, b, 0): b \in \mathbb{N}\} \cup\{(a, b, 1): a, b \in \mathbb{N}\}$
First, we recognize that the right-hand side is factored:

$$
a^{b+20}(c-1)=\left(c^{b+20}+c^{b+19}+\cdots+c+1\right)(c-1) .
$$

The case $c=1$ is treated separately (and is very simple). For $c \neq 1$, the equation simplifies to

$$
a^{b+20}=c^{b+20}+c^{b+19}+\cdots+c+1 .
$$

We therefore distinguish the two cases for $c$.

- $c=1$ leads (see above) to

$$
0=0 \text {. }
$$

Therefore, any natural numbers for $a$ and $b$ are solutions.

- For $c \neq 1$, we can (see above) divide by $c-1$ and obtain the equivalent equation

$$
a^{b+20}=c^{b+20}+c^{b+19}+\cdots+c+1 .
$$

Obviously,

$$
c^{b+20}+c^{b+19}+\cdots+c+1>c^{b+20} .
$$

Therefore, $a \geq c+1$ must hold. By the binomial theorem,

$$
\begin{aligned}
a^{b+20} & \geq(c+1)^{b+20} \\
& =c^{b+20}+\left(\begin{array}{c}
b+20 \\
1
\end{array}\right) c^{b+19}+\cdots+\left(\begin{array}{l}
b+20 \\
b+19
\end{array}\right) c+1 \\
& \geq c^{b+20}+c^{b+19}+\cdots+c+1 \\
& =a^{b+20} .
\end{aligned}
$$

Therefore, both inequalities must actually be equalities.

We particularly consider the second inequality. Since

$$
\left(\begin{array}{c}
b+20 \\
1
\end{array}\right)=b+20>1
$$

this can only be an equality if $c=0$. In the case $c>0$, the second inequality is strict and thus leads to a contradiction, and there is no solution.

In the remaining case $c=0$, the resulting equation is

$$
a^{b+20}=1
$$

which is easy to solve. Since $a$ is a natural number, $a=1$ must hold. (This also follows from the necessary relationship $a=c+1$.) Therefore, $b$ can be any natural number.

Alternatively, under the assumption $c>0$, we recognize directly that

$$
\begin{aligned}
c^{b+20} & <c^{b+20}+c^{b+19}+\cdots+c+1 \\
& <c^{b+20}+\left(\begin{array}{c}
b+20 \\
1
\end{array}\right) c^{b+19}+\cdots+\left(\begin{array}{l}
b+20 \\
b+19
\end{array}\right) c+1 \\
& =(c+1)^{b+20}
\end{aligned}
$$

holds. Therefore, $a^{b+20}$ lies strictly between $c^{b+20}$ and $(c+1)^{b+20}$, which is impossible for natural numbers. (The case $c=0$ must then be treated separately as above.)

(Michael Drmota)