The only solutions in positive integers of the Diophantine equation $(1) \;\; k + a^k = m + 2a^m$ are $(a,k,m) = (1,t+1,t)$, where $t$ is a positive integer. Proof: Equation (1) is equivalent to $(2) \;\; k - m = 2a^m - a^k$. If $m \geq k$, then $k - m \leq 0$ and $2a^m - a^k \geq 2a^k - a^k = a^k$, contradicting equation (2). Consequently $k > m$, yielding $a^m \mid k - m$ by equation (2). Hence by equation (2) ${\textstyle 0 < \frac{k - m}{a^m} = 2 - a^{k - m}}$, yielding $a^{k-m}<2$. Therefore $a^{k-m}=1$, which means $a=1$ (since $k-m \geq 1$), which inserted in equation (1) give us $m=k+1$. Summa summarum, the only solutions in positive integers of equation (1) are $(a,k,m) = (1,t+1,t)$, where $t$ is a positive integer. q.e.d.