Mathematical Reflections 2016, Issue 5 - Problem O386

Problem:
Find all pairs $(m,n)$ of positive integers such that $3^m-2^n$ is a perfect square.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Let $n \in \mathbb{N}$ such that
$$3^x-2^y=n^2 \qquad (1).$$
We have three cases.

(i) If $y \geq 2$, then $3^x \equiv n^2 \pmod{4}$, so $x$ must be even, i.e. $x=2k$ for some $k \in \mathbb{N}^*$. Therefore, equation (1) becomes $$(3^k-n)(3^k+n)=2^y.$$ It follows that $3^k-n=2^a$ and $3^k+n=2^b$, where $a,b \in \mathbb{N}$ and $a+b=y$. Moreover, $a<b$ and adding these two equations, we get $$2\cdot3^k=2^a+2^b.$$ If $a=0$, the LHS is even and the RHS is odd, contradiction. If $a \geq 2$, we obtain $2\cdot3^k \equiv 0 \pmod{4}$, contradiction. So, $a=1$ and $b=y-1$ and we get $$3^k=1+2^{y-2}.$$ If $y=3$, we obtain $k=1$, i.e. $x=2$. If $y \geq 4$, then $3^k \equiv 1 \pmod{4}$, so $k$ is even and we can write $$(3^{\frac{k}{2}}-1)(3^{\frac{k}{2}}+1)=2^{y-2}.$$ Since $3^{\frac{k}{2}}+1$ and $3^{\frac{k}{2}}-1$ are powers of $2$ and their difference is $2$, we obtain $3^{\frac{k}{2}}+1=4$ and $3^{\frac{k}{2}}-1=2$, i.e. $k=2$, which gives $x=4$ and $y=5$. Therefore, we obtain the solutions $(x,y) \in \{(2,3),(4,5)\}$.

(ii) If $y=1$, then $3^x-2=n^2$. We have that $x$ must be odd, otherwise $n^2 \equiv -1 \pmod{4}$, contradiction. In the ring of integers $\mathbb{Z}[\sqrt{-2}]$, we have $$3^x=(n-\sqrt{-2})(n+\sqrt{-2}).$$
Let $d=(n-\sqrt{-2},n+\sqrt{-2})$. Clearly, $d \ | \ 2\sqrt{-2}$, so $N(d) \ | \ 8$. Since $3^x$ is odd, then its norm is odd and this implies that $N(d)=1$, i.e. $d=\pm 1$. So, $n-\sqrt{-2}$ and $n+\sqrt{-2}$ are coprime in $\mathbb{Z}[\sqrt{-2}]$. Since these two factors have the same norm and the only non trivial factorization (up to sign permutations) of $3$ with factors with the same norm is $3=(1-\sqrt{-2})(1+\sqrt{-2})$, then this forces
$$\begin{array}{rcl} n-\sqrt{-2}&=&(1-\sqrt{-2})^x \\ n+\sqrt{-2}&=&(1+\sqrt{-2})^x  \end{array} \qquad \textrm{or} \qquad \begin{array}{rcl} n-\sqrt{-2}&=&(1+\sqrt{-2})^x \\ n+\sqrt{-2}&=&(1-\sqrt{-2})^x  \end{array}$$
Considering the first equation, we get $\textrm{Im}(n-\sqrt{-2})=\textrm{Im}((1-\sqrt{-2})^x)$, i.e. $$-\sqrt{2}=\left[-{x \choose 1}+2{x \choose 3}-4{x \choose 5}+\ldots+(-1)^{\frac{x+1}{2}}2^{\frac{x-1}{2}}{x \choose x}\right]\sqrt{2},$$ i.e. $$-1=-{x \choose 1}+2{x \choose 3}-4{x \choose 5}+\ldots+(-1)^{\frac{x+1}{2}}2^{\frac{x-1}{2}}{x \choose x}.$$ Let $$\displaystyle f(x)=\sum_{k \textrm{ odd}}^x {x \choose k} (-1)^{\frac{k+1}{2}}2^{\frac{k-1}{2}}$$ be defined on the odd natural numbers. An easy check shows that $f(x) \leq f(x+2)$ for any odd $x$. Since $f(5)=11$, then $x \in \{1,3\}$. An easy check shows that $f(1)=f(3)=-1$ and we get $(x,y) \in \{(1,1),(3,1)\}$. If we consider the second system of equations, we get $\textrm{Im}(n-\sqrt{-2})=\textrm{Im}((1+\sqrt{-2})^x)$ and proceeding as before, we obtain no solutions.

(iii) If $y=0$, then $3^x-1=n^2$, i.e. $3^x=n^2+1$. Since $x>0$ and $n^2 \equiv 0,1 \pmod{3}$, we get no solutions in this case.

In conclusion, $(x,y) \in \{(1,1),(2,3),(3,1),(4,5)\}$.