Power Diophantine equation: $(a+1)^n=a^{n+2}+(2a+1)^{n-1}$

Question

How to solve following power Diophantine equation in positive integers with $n>1$:$$(a+1)^n=a^{n+2}+(2a+1)^{n-1}$$

Answer 1

If the equation holds, then $(a+1)^n>(2a+1)^{n-1}$ and $(a+1)^n>a^{n+2}$.

Multiplying gives $(a+1)^{2n} > (2a+1)^{n-1}a^{n+2}$. This can be rewritten as $(a^2+2a+1)^n > (2a^2+a)^{n-1} a^3$. However, if $a\geq 3$, then $a^2+2a+1 < 2a^2+a$ and $a^2+2a+1 <a^3$, which gives a contradiction.

Hence $a=1$ or $a=2$. The first gives $2^n=1+3^{n-1}$. With induction, one may prove that the right hand side is larger for $n>2$. This only gives the solution $(1,2)$.

The second gives $3^n=2^{n+2}+5^{n-1}$. With induction, one may prove that the right hand side is larger for $n>3$. One can also check that there is no equality for $n=1,2,3$.