Mathematical Reflections 2017, Issue 1 - Problem J401

Problem:
Find all integers $n$ for which $n^2+2^n$ is a perfect square.

Proposed by Adrian Andreescu, Dallas, Texas

Solution:
Clearly, $n \geq 0$. If $n=0$, we get $n^2+2^n=1$, which is a perfect square. Let $n>0$. If $n$ is even, then $n=2k$ for some $k \in \mathbb{N}^*$. If $k \geq 7$, then $$(2^k)^2=2^{2k}<4k^2+2^{2k}<2^{2k}+2^{k+1}+1=(2^k+1)^2,$$ so $n^2+2^n$ is not a perfect square if $n$ is even and $n \geq 14$. So, $n \in \{2,4,6,8,10,12\}$. An easy check gives the solution $n=6$. If $n$ is odd, then $n=2k+1$ for some $k \in \mathbb{N}$. If $k=0$, we get no solutions, so assume $k \geq 1$. Let $m \in \mathbb{N}^*$ such that $(2k+1)^2+2^{2k+1}=m^2$. Then, $$(m-2k-1)(m+2k+1)=2^{2k+1}.$$ Since $m-2k-1<m+2k+1$ and the two factors have the same parity, then $$\begin{array}{lll} m-2k-1&=&2^a \\ m+2k+1&=&2^b, \end{array}$$ where $a,b \in \mathbb{N}$, $1 \leq a \leq b \leq 2k$ and $a+b=2k+1$. If $a \geq 2$, then subtracting we get $2(2k+1)=2^b-2^a=2^a(2^{b-a}-1)$, i.e. $2k+1=2^{a-1}(2^{b-a}-1)$, contradiction. So, $a=1$ and $b=2k$, which gives $2k+1=2^{2k-1}-1$, i.e. $k=2^{2k-2}-1$. If $k \geq 2$, then $k<2^{2k-2}-1$, so it must be $k=1$. But if $k=1$, we get no solutions. So, there are no solutions when $n$ is odd. We conclude that $n \in \{0,6\}$.