Problem 1543

Solution:

If we expand $(a+1)^{n}+a-1$ using the binomial theorem, we get:
$$a^{n}+\cdots+\frac{n(n-1)}{2} a^{2}+n a+1+a-1=a^{n}+\cdots+\frac{n(n-1)}{2} a^{2}+(n+1) a.$$
Thus, since all the terms are divisible by $a$, and since $(a+1)^{n}+a-1$ is a power of $2$, $a$ is also a power of $2$. Let us call $a=2^{b}$ and $(a+1)^{n}+a-1=2^{c}$. From the condition $a \geq 2$ we obtain $b \geq 1$. Moreover, from the condition $n \geq 2$ we also obtain that $2^{c}>a^{2}=2^{2b}$, and hence $c>2b$.
Observe that all the terms in $(\star)$, except possibly the last one, are divisible by $a^{2}=2^{2b}$.
Since $c>2b, 2^{c}$ is divisible by $2^{2b}$, and hence, by difference, so is $(n+1) a$. Since $a=2^{b}$, it follows that $2^{b}$ divides $n+1$, that is, $n+1=2^{b} \cdot m=a m$ for some positive integer $m$. From the condition $a \geq n \geq 2$, the only possible value for $m$ is $m=1$, and hence $n=a-1=2^{b}-1$. In particular, $b$ cannot take the value $1$, otherwise $n=1$, and hence $b>1$, from which $a \geq 4$ and $n=a-1 \geq 3$. From this last inequality it follows that $2^{c}>a^{3}$, and hence $c>3b$.
Let us rewrite $(a+1)^{n}+a-1$ using the information we have gathered:
$$\begin{aligned} (a+1)^{n}+a-1 & =a^{n}+\cdots+\frac{n(n-1)(n-2)}{6} a^{3}+\frac{n(n-1)}{2} a^{2}+(n+1) a= \\ & =2^{nb}+\cdots+\frac{\left(2^{b}-1\right)\left(2^{b}-2\right)\left(2^{b}-3\right)}{6} 2^{3b}+\frac{\left(2^{b}-1\right)\left(2^{b}-2\right)}{2} 2^{2b}+2^{2b} \end{aligned}$$
All the terms, except possibly the last two, are divisible by $2^{3b}$; moreover $2^{c}$ is also divisible by $2^{3b}$, and hence so is $\frac{\left(2^{b}-1\right)\left(2^{b}-2\right)}{2} 2^{2b}+2^{2b}=\left(2^{b}-1\right)\left(2^{b-1}-1\right) 2^{2b}+2^{2b}$.
It follows that $\left(2^{b}-1\right)\left(2^{b-1}-1\right)+1=2^{2b-1}-2^{b}-2^{b-1}+2$ is divisible by $2^{b}$, but this is possible only when $b=2$: if $b>2$, in fact, all the terms of the expression except the last one are divisible by $4$, and hence the sum is not divisible by $4$ (and hence not by $2^{b}$ either).
The only remaining case is $b=2$, from which $a=4$ and $n=3$. In this case, an elementary check shows that $(a+1)^{n}+a-1=128=2^{7}$, so this is a solution.
In conclusion, there is a unique solution: $a=4, n=3$.