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Determine all pairs of perfect powers $2^m$, $3^n$ such that their difference is a perfect square.
I found these pairs work by mod 3 and mode 4:
Case 1, $2^m \ge 3^n$, $(0,0), (2,1), (1,0)$
Case 2, $2^m \le 3^n$, $(0,0), (1,1), (1,3), (3, 2), (5, 4)$
How to prove there's no others?
I also found this post: Pairs forming perfect square. There's good discussions. Do we have any complete solution?
You can solve the problem in three cases:
Case 1: $m\ge2$ and $2^m>3^n$.
Here, looking modulos $3$ and $4$ demands that $m$ is even and $n$ is odd. The problem then reduces to the form $$(2^{m/2})^2=x^2+3(3^{(n-1)/2})^2\,~\,~\,~\,~ (1)$$ for some integer $x$; thus we are looking at solutions to the Diophantine equation $$z^2=x^2+3y^2\,.$$ Checking modulo $8$ shows that the only possibility for both $x,y$ to be odd is when $z^2\equiv 4\mod 8$, which thus demands that the only solutions to Eq (1) must be $$(m,n)=(2,1)\,.$$
Case 2: $m\ge2$ and $2^m<3^n$.
Here also, looking modulos $3$ and $4$ rather demands that $m$ is odd and $n$ is even. Now, the problem reduces to the form $$(3^{n/2})^2=x^2+2(2^{(m-1)/2})^2\,~\,~\,~\,~ (2)$$ for some integer $x$; thus, this time, we are looking at solutions to the Diophantine equation $$z^2=x^2+2y^2\,.$$ The general solution to this equation (in the manner similar to deriving Pythagorean Triples) is given by $$z=a^2+2b^2\,,~\,~\,~\,x=a^2-2b^2\,,~\,~\,~\,~y=2ab$$ for some integers $a,b$. Comparing to Eq (2) forces $a=\pm 1$ and $b=\pm 2^{(m-3)/2}$, which implies that $3^{n/2}=z=(\pm 1)^2+2(\pm 2^{(m-3)/2})^2= 1+2^{m-2}$——that is, $3^{n/2}-2^{m-2}=1$——which has the only solutions $$(m,n)\in\{(3,2),(5,4)\}$$ thanks Catalan-Mihailescu’s Theorem.
Case: $m\in\{0,1\}$.
For $n=0$, we have the obvious solutions $$(m,n)\in\{0,0),(1,0)\}\,.$$ Hence suppose $n\ge 1$. Here, we have $3^n-2^m=y^2$ for some integer $y$, which can be rewritten as $$y^2=3^{e_n}\left(3^{(n-e_n)/3)}\right)^3-2^m\,,$$ where $e_n\in\{0,1,2\}$ and $n\equiv e_n\mod 3$,. Multiplying through by $3^{2e_n}$, we see that we are actually looking at integer solutions to the Mordell elliptic curve $$y^2=x^3-2^m\cdot 3^{2e_n}\,,$$ and a solution to your problem corresponds to $x=3^{(n+2e_n)/3}$. Mordell (1920) proved that there are only finitely many integral solutions to such elliptic curves (you can find the integer solutions listed here https://web.archive.org/web/20110618033146/http://tnt.math.se.tmu.ac.jp/simath/MORDELL/MORDELL-); in particular, for $m=1$ and $e_n=0$, it is a theorem or Fermat that the only integral solutions to the elliptic are $(x,y)\in\{3,\pm 5\}$, which thus gives the only solution to your problem as $$(m,n)=(1,3)\,.$$ When $m=e_n=0$, the only integral solution is $(x,y)=(1,0)$, which doesn’t correspond to a solution to your problem. For $e_n=1$, using the linked address above, the only integral solutions are no solutions for $m=0$ but $(x,y)=(3,\pm 3)$ for $m=1$, which forces the only solution to your problem in this case $$(m,n)=(1,1)\,.$$ Finally, for $e_n=2$, there are no integral solutions to the elliptic curve for $m=1$, but for $m=0$, the only solution is $(x,y)=(13,46)$, which does not correspond to a solution to your problem.
This is a partial solution concerning $m\ne 1$.
First we consider the cases $m\ge 2$.
For case 1 ($2^m \ge 3^n$), write $2^m = 3^n + k^2$.
Taking modulo $3$, we have $(-1)^m \equiv k^2 \pmod 3$. This gives $m$ is even.
Write $m = 2s$, then $3^n = 2^m-k^2 = (2^s-k)(2^s+k)$, and both $2^s \pm k$ are powers of $3$.
Summing them up, however, gives $2^{s+1}$ which is not a multiple of $3$. Hence $2^s-k = 1$.
$2^s+k = 2^{s+1}-1$ is a power of $3$. [This gives $s=1, m=2, n=1$.]
For case 2 ($2^m \le 3^n$), write $3^n = 2^m + k^2$.
Taking modulo $4$, we have $(-1)^n\equiv k^2 \pmod 4$. This gives $n$ is even.
Write $n=2t$, then $2^m = 3^n-k^2 = (3^t-k)(3^t+k)$, and both $3^t \pm k$ are powers of $2$.
If we write $3^t-k = 2^x, 3^t+k =2^y$, we have $2^x+2^y = 2^x(1+2^{y-x})$.
Summing them up, however, gives $2 \times 3^t$. Since $3^t$ is odd, we must have $2^x=2$ and $1+2^{y-1} = 3^t$.
[This equation has $2$ solutions: $y=2, t=1$ and $y=4, t=2$.]
Both cases above end with solving an equation of the form $1+a^b= c^d$, which by Catalan's Conjecture/Mihăilescu's theorem only has the solution $a=2,b=3,c=3,d=2$, for integers $a,b,c,d >1$. However there are elementary proofs for this special case $a,c \in \{2,3\}$, which I choose to omit (unless requested)
For the case $m=0$, write $3^n = 1+k^2$. By modulo $3$, no solutions exist for $n \ge 1$. This forces $n=0$.
For the case $m=1$, write $3^n = 2+k^2$. By modulo $3$ we have $k^2 \equiv 1\pmod 3$, and by modulo $4$ we have $(-1)^n \equiv 2+k^2$, forcing $n$ to be odd and $k^2\equiv 1 \pmod 4$. $k$ must be an odd number not divisible by $3$, but as of right now I am not sure how to proceed.
