Mathematical Reflections 2017, Issue 1 - Problem J397

Problem:
Find all positive integers $n$ for which $3^4+3^5+3^6+3^7+3^n$ is a perfect square.

Proposed by Adrian Andreescu, Dallas, Texas

Solution:
We have $3^4+3^5+3^6+3^7+3^n=3240+3^n$. If $n \leq 4$, we obtain that $3240+3^n$ is a perfect square only when $n=2$. Assume that $n>4$. Then, there exists $t \in \mathbb{N}$ such that $3240+3^n=t^2$, i.e. $$3^4(40+3^{n-4})=t^2.$$ Since $3^4$ is a perfect square, it follows that $40+3^{n-4}$ must be a perfect square, so there exists $m \in \mathbb{N}$ such that $40+3^{n-4}=m^2$. If $n-4$ is odd, then $40+3^{n-4} \equiv 3 \pmod{4}$, so it cannot be a perfect square. It follows that $n-4$ is even, i.e. $n-4=2k$, where $k \in \mathbb{N}$, $k>2$. So, $40+3^{2k}=m^2$, i.e. $$(m-3^k)(m+3^k)=40.$$ Since $m-3^k<m+3^k$ and both factors have the same parity, then we obtain the two systems of equations
$$\begin{array}{lll} m-3^k&=&2 \\ m+3^k&=&20, \end{array} \qquad \begin{array}{lll} m-3^k&=&4 \\ m+3^k&=&10. \end{array}$$
Subtracting the first equation to the second equation of each system, we get $3^k=9$ or $3^k=3$, i.e. $k=2$ or $k=1$. So, $n=8$ or $n=6$. We conclude that $n \in \{2,6,8\}$.