I’m working on a problem, and believe I have reduced it to the solution of the Diophantine equation $$23^n+1=2x^2.$$
Brute force calculations suggest that the only solutions are $(n,x)=(0,1)$ and $(n,x)=(3,78)$. I haven’t proven it algebraically yet, but wanted to know if it’s a known result?
On
COMMENT.- Maybe the following can be useful: The solution of $2x^2\equiv1\pmod{23}$ is $x\equiv\pm9\pmod{23}$ so one has in $\mathbb Z$ the equations $$23^{n-1}=7+36m+23m^2\\23^{n-1}=17+56m+23m^2$$ from which $$n-1=\frac{\log(23m^2+36m+7)}{\log(23)}\\n-1=\frac{\log(23m^2+56m+17)}{\log(23)}$$ it remains to show that $n$ cannot be an integer other than $0$ and $3$.
On
as you noticed, $n=0$ gives $x = \pm 1$.
if $n$ is even, $n=2m >0$, we have:
$$23^{2m} +1 = 2x^2 \iff (23^{m})^2-2x^2 = -1$$
which is a generalized Pell's equation, whose candidates to solutions are of the form $a+b\sqrt2$. In particular: $$y^2-2x^2 = -1 $$
You can check here that the general solution for that equation is $$y_a + x_a \sqrt 2 = (1+\sqrt2)^{2a-1}, a \in \mathbb N$$
we want one $y_a$ such that $y_a = 23^m $. I hope you will agree with me that $$y_a = \frac{(1+\sqrt2)^{2a-1} + (1-\sqrt2)^{2a-1}}{2}$$ Now we use a result from the non-trivial-by-any-means Carmichael's theorem. Notice that we are dealing with a subset of the Pell's numbers and the wikipedia link that I posted says that every Pell's number other than $1$ is such that it has new prime factors that the previous ones didn't have, but $y_6 = 8119 = 23 \cdot 353$ is the first one divisible by $23$ so any other $y_a$ will have factors other than $23$.
So $n$ must be odd. Now look:
$$23^{2b+1} +1 = 2x^2 \iff 2x^2 - 23 \cdot 23^{2b} = 1$$
we can try to apply the same principle here, but the theorem I used is very strong and particular. It is easy to find the general formula for the solutions of the Pell's equation $$2x^2-23y^2 =1$$, but maybe you guys have a better idea to solve the odd case. Good luck :D
Comments already contain enough information to craft an answer.
Inspecting both sides mod $7$ we find that only possible remainder is $2$, i.e. $$ 23^n+1\equiv 2x^2\equiv 2\pmod 7. $$ Since $23^n\equiv 1\pmod 7$ is equivalent to $3\mid n$, we can assume $n=3k$ for some integer $k$. Then the original equation becomes $$ (23^{k})^3+1=2x^2. $$ Multiplying both sides by $8$ and letting $a=2\cdot23^k$ and $b=4x$, we have $$ a^3+8=b^2. $$ This is a Mordell curve with finitely many solutions $(a,b)$ being $(−2, 0), (1, 3), (2, 4), (46, 312)$. Only $(2,4)$ and $(46,312)$ translate to the solution of the original problem, giving $n=0,x=1$ and $n=3,x=78$ respectively.