SELECTION EXAMINATION

Since $a$ is an even positive integer, it follows that $A$ is odd. Therefore $A$ will be a perfect square of an odd integer, that is $$A=(2\kappa +1{)}^2=4{\kappa }^2+4\kappa +1=4\kappa (\kappa +1)+1,$$ where $\kappa$ is a positive integer. Since one of the two integers $\kappa$ and $\kappa +1$ is even, we have $$\begin{array}{rl}A& =4\kappa (\kappa +1)+1=8\rho +1,\text{ where }\rho \text{ is a positive integer}\\ \Rightarrow A-1& =a^n+a^{n-1}+\cdots +a=8\rho \Rightarrow a(a^{n-1}+\cdots +a+1)=8\rho \\ \Rightarrow 8|a(a^{n-1}+\cdots +a+1)\Rightarrow 8|a,\text{ since }(8,a^{n-1}+\cdots +a+1)=1.& \end{array}$$