10

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?

Guruprasad
  • 803
Kieren MacMillan
  • 8,199
  • what were your bounds? I would say try for $n \simeq \frac{2 \log x + \log 2}{\log 23}$ so the $x$ must be way greater than $n$ – julio_es_sui_glace Sep 09 '23 at 21:22
  • @julio_es_sui_glace: I’m currently at $n=2300$ and climbing — not massive yet, I know, but I’m working on it! – Kieren MacMillan Sep 09 '23 at 21:24
  • Ok still impressive though, good luck (+1) – julio_es_sui_glace Sep 09 '23 at 21:25
  • 2
    Note that if $n$ is odd, then it must be a multiple of $3$. This is due to $23^n+1=(23+1)(23^{n-1}-23^{n-2}+\ldots -23+1)$. With $23\equiv -1\pmod{3}$, so $23^{n-1}-23^{n-2}+\ldots -23+1 \equiv 1+1+\ldots+1+1\equiv n\pmod{3}$. Thus, Since as $23+1=3(2^3)$, then $3\mid x$, so there must be at least $2$ factors of $3$, we have $3 \mid n$. Also, FYI, this could be determined using the Lifting-the-exponent lemma, which actually shows more specifically that $n$ must have an odd # of factors of $3$. – John Omielan Sep 09 '23 at 21:46
  • 2
    Furthermore, for odd $n$, with $3\mid n$, then also $23^{3}+1\mid 23^n+1$. Since $23^3+1=2^3(3^2)(13^2)$, the LTE lemma then shows that $n$ must have an even number of factors of $13$. – John Omielan Sep 09 '23 at 22:19
  • Checked up to n=500000 with Sage and no further solutions found (assuming I didn't make any mistakes in my code). – John Palmieri Sep 09 '23 at 23:23
  • We take the three cases $n=3a, n=3a+1,$ and $n=3a+2$. The problem can be reduced to finding the integer points on elliptic curves as follows. $Y^2=X^3+8, Y^2=X^3+4232, Y^2=X^3+2238728$. – Tomita Sep 10 '23 at 00:17
  • @Tomita: according to Sage, the 2nd and 3rd elliptic curves have no integral points. The first has four: (-2 : 0 : 1), (1 : -3 : 1), (2 : -4 : 1), (46 : -312 : 1). – John Palmieri Sep 10 '23 at 00:21
  • 2
    $23^n+1\equiv 2,3,5 \pmod 7 $ and $2x^2\equiv 0,1,2,4\pmod 7$, so $23^n+1\equiv 2x^2\equiv 2\pmod 7$. This too implies $3\mid n$. – Sil Sep 10 '23 at 00:29
  • @JohnOmielan: Is there an easy proof that $n=0$ is the only even solution? – Kieren MacMillan Sep 10 '23 at 01:06
  • According to John Palmieri, $(X,Y)=(2,4) \implies a=0, (X,Y)=(46,312) \implies a=1$. – Tomita Sep 10 '23 at 02:08

3 Answers3

9

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.

Sil
  • 17,976
3

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

hellofriends
  • 2,048
2

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 $3$.

Ataulfo
  • 32,657