Proving that there are no other integer solutions

Question

The integer solutions for this equation: $m ^ 2 = 2 \cdot 3 ^ n - 5$ are these four couples: $(n; m): (1;-1) (1;1) (3;7) (3;-7)$.

How can I prove there are no other solutions?

Answer 1

One mechanical way is to reduce this to a finite number of Mordell curves: $3^n$ always belongs to one of the forms $r^3$, $3r^3$, or $9r^3$. Doing a simple change of variables transforms each of these possibilities into an equation of the form $y^2 = x^3 - N$, where $N \in \{20, 180, 1620\}$. There are precomputed tables of solutions to these curves, and while OEIS doesn't carry the solutions themselves, it does tabulate the number of integer points at http://oeis.org/A081120:

• $N=20$ has $2$ points, so they must be $(x,y) = (6,\pm 14)$ corresponding to your solutions $(n,m) = (3,\pm 7)$ after unravelling the change of variables.
• $N=180$ has $2$ points, so they must be $(x,y) = (6,\pm 6)$ corresponding to $(n,m) = (1,\pm 1)$.
• $N=1620$ has no integer points, so all solutions are accounted for.

Edit: In the comments, Peter points out that there is a significant error in the OEIS listing for $N=180$ (which likely indicates more errors). There is an additional solution $(x,y) = (69,\pm 573)$, but this leads to a half-integer value of $m = \frac{191}{2}$ and an irrational value for $n$, so it doesn't contribute a new solution to the original equation.

Answer 2

An even $n$ cannot lead to a solution because the equation can be rewritten as $$m^2-2k^2=-5$$ since $3^n$ is a square, but this equation has no solution modulo $25$, hence no solutions in the integers.

If $n$ is odd and $n\ge 5$, the equation can be rewritten as $$m^2-486k^2=-5$$

The convergents $\frac{p_n}{q_n}$ of the continued fraction of $\sqrt{486}$ satisfy $$p_n^2-486q_n^2=-2$$ for odd n and $$p_n^2-486q_n^2=1$$ for even $n$

Since $|-5|=5<\sqrt{486}$ and $-5$ does not occur in the sequence $p_n^2-486q_n^2$, we can conclude that $m^2-486k^2=-5$ has no integer solution. This completes the proof that there are no solutions for $n\ge 5$

Answer 3

Here is an approach.

Since $m$ is odd, let $m = 2k + 1$

We have $4k^2 + 4k + 6 = 2*3^n \implies 2k^2 + 2k + 3 = 3^n \implies 2k(k+1) = 3(3^{n-1} - 1)$

So $3$ has to divide $k$ or $k+1$ and $3$ cannot be a factor of both $k$ and $k+1$.

If $3$ divides $k$, then $k$ divides $3 \implies k = 3 \implies m = 7, n = 3$

If $3$ divides $k$ and $k+1$ divides $3 \implies k = 0, m = 1, n = 1$