Austrian Mathematical Olympiad

Answer. $\{(1,b,0):b\in {\mathbb{Z}}_{>0}\}\cup \{(a,b,1):a,b\in {\mathbb{Z}}_{>0}\}$

One can first see that the right side factors: $$a^{b+20}(c-1)=(c^{b+20}+c^{b+19}+\cdots +c+1)(c-1).$$ The case $c=1$ will be handled separately (and is very simple). For $c\ne 1$ the equation simplifies to $$a^{b+20}=c^{b+20}+c^{b+19}+\cdots +c+1.$$

* For $c\ne 1$ we can divide by $c-1$ (see the above equations) and get 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 $a\ge c+1$ must hold. Because of the binomial theorem we, thus, obtain $$\begin{array}{rl}{a}^{b+20}& \ge (c+1{)}^{b+20}\\ & =c^{b+20}+\left(\genfrac{}{}{0ex}{}{b+20}{1}\right)c^{b+19}+\cdots +\left(\genfrac{}{}{0ex}{}{b+20}{b+19}\right)c+1\\ & \ge c^{b+20}+c^{b+19}+\cdots +c+1\\ & =a^{b+20}.\end{array}$$ Hence, both inequalities must be equations. We consider the second inequality in particular. Because of $$\left(\genfrac{}{}{0ex}{}{b+20}{1}\right)=b+20>1$$ this can only be an equation if $c=0$. In the case of $c>0$, the second inequality is strict and therefore leads to a contradiction and there is no solution. In the remaining case $c=0$, the resulting equation $$a^{b+20}=1$$ is easy to solve. Since $a$ is a natural number, $a=1$. (This also follows from the necessary relationship $a=c+1$.) Hence, in this case $b$ may be any natural number. Alternatively, one can see in the case $c>0$ that $$\begin{array}{rl}{c}^{b+20}& <c^{b+20}+c^{b+19}+\cdots +c+1\\ & <c^{b+20}+\left(\genfrac{}{}{0ex}{}{b+20}{1}\right)c^{b+19}+\cdots +\left(\genfrac{}{}{0ex}{}{b+20}{b+19}\right)c+1\\ & =(c+1{)}^{b+20}.\end{array}$$ So $a^{b+20}$ is in this case 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.)