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I know that $n$ can't be even because of the following argument:

Let $n = 2p$. Then we can use the difference of two squares and it becomes like this : $(3^p + x)(3^p - x) = 11; 3^p + x = 11 , 3^p - x = 1$. $3^p = 6$ which is not possible if $p$ is an integer.

I also found out that $x$ has to be an even number. I think that the only solution is $3^3 - 4^2 = 11$ but how can I find the true answer?

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    $\begingroup$ $\mathbb{Z}[\frac{1 + \sqrt{-11}}{2}]$ is a Euclidean domain. $\endgroup$
    – user44191
    Commented Aug 16, 2021 at 14:36
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    $\begingroup$ Equations of this type are called Ramanujan-Nagell type. This term might help find some relevant tools. $\endgroup$
    – Wojowu
    Commented Aug 16, 2021 at 14:45
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    $\begingroup$ This particular equation is considered in papers Alter & Kubota (1975) and Cohen (1976). $\endgroup$
    – Max Alekseyev
    Commented Aug 17, 2021 at 18:37
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    $\begingroup$ Just for the record, a similar question: mathoverflow.net/q/384484 $\endgroup$
    – Max Alekseyev
    Commented Aug 18, 2021 at 14:33

2 Answers 2

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One way to do this (probably not the easiest, but generalizes quite nicely) is to write $n=3m+r$ with $r\in\{0,1,2\}$. Then you have a solution $(x,3^m)\in\mathbb Z^2$ to $$ 3^r\cdot y^3 - x^2 = 11. $$ So you just need to find the integer points on the three elliptic curves $$ y^3-x^2=11,\quad 3y^3-x^2=11,\quad 9y^3-x^2=11. $$ There are standard methods for handling such curves.

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    $\begingroup$ Is there any particular reason to write $n = 3m + r$ rather than $n = 2m + r$? $\endgroup$
    – user44191
    Commented Aug 16, 2021 at 15:08
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    $\begingroup$ @user44191 An equation $Q(x,y)=c$ with $Q$ quadratic may have infinitely many integer solutions, e.g., $x^2-Dy^2=1$. So after finding a generator for the infinitely many solutions, you'd still need to figure out which ones have their $y$-coordinate a power of 3. OTOH, an elliptic curve always has finitely many integer points; and there is in principle an effective algorithm to find them. In practice, there may be computational issues; and cleverness never hurts. I'm not sure to what extent this can be done automatically for small coefficients, e.g., in Sage or other packages. $\endgroup$
    – Joe Silverman
    Commented Aug 16, 2021 at 15:24
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    $\begingroup$ These are further reduced to Mordell equations: \begin{split} x^2&=y^3-11\\ (3x)^2 &= (3y)^3 - 99 \\ (9x)^2 &= (9y)^3 - 891\end{split} with readily available solutions. $\endgroup$
    – Max Alekseyev
    Commented Aug 16, 2021 at 23:55
  • $\begingroup$ @MaxAlekseyev Thanks for providing the link to solve this particular case, but since the OP didn't say why this particular equation is interesting (it doesn't look so to me), I took the real question to be "how does one solve such problems." If the numbers are small enough, one can look up the answers in the link you cite (or elsewhere). If the numbers are fairly small, there are packages that will find the solutions (sometimes rigorously, sometimes with some heuristics), and if the numbers are quite large, then one may be out of luck for the present. $\endgroup$
    – Joe Silverman
    Commented Aug 17, 2021 at 1:22
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    $\begingroup$ If one wanted to avoid MordelI equations, I guess one could use something like $|3^n -x^2| > 3^{7n/40}$, valid for all integer $x$ and odd $n$, to solve such equations, even with reasonable large numbers involved. $\endgroup$
    – Mike Bennett
    Commented Aug 19, 2021 at 21:51
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$$3^{n} - x^2 = 11$$

According to Silverman's answer, 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.

$\bullet n=3a$
Let $X = 3^a, Y=x$, then we get $Y^2 = X^3 - 11.$
According to LMFDB, this elliptic curve has integer points $(X,Y)=(3,\pm 4), (15,\pm 58)$ with rank $2.$
Hence $(X,Y)= (3,\pm 4) \implies (n,x)=(3,\pm4).$

$\bullet n=3a+1$
Let $X = 3^{a+1}, Y=3x$, then we get $Y^2 = X^3 - 99.$
This elliptic curve has rank $0$ and has no integer point, so there is no integer solution $(n,x).$

$\bullet n=3a+2$
Let $X = 3^{a+2}, Y=9x$, then we get $Y^2 = X^3 - 891.$
This elliptic curve has integer points $(X,Y)=(31,\pm 170)$ with rank $1.$
Hence there is no integer solution $(n,x).$

Thus, there are only integer solutions $(n,x)=(3,\pm4).$

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  • $\begingroup$ As I mentioned in the other comment, these are Mordell equations with readily available solutions. $\endgroup$
    – Max Alekseyev
    Commented Aug 16, 2021 at 23:58
  • $\begingroup$ @Max Alekseyev, Thank you for your information. $\endgroup$
    – Tomita
    Commented Aug 17, 2021 at 0:38
  • $\begingroup$ Uff! This simple angle of view gives a really astonishing blink of insight... $\endgroup$
    – Gottfried Helms
    Commented Aug 17, 2021 at 9:27
  • $\begingroup$ @Gottfried Helms, I just applied Silverman's method to the problem. Thanks. $\endgroup$
    – Tomita
    Commented Aug 17, 2021 at 10:07
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    $\begingroup$ @Tomita I appreciate the kudos, but if you ever refer to it in an article or such, you should not call it "Silverman's method." It's a standard and well-known technique that I'm sure was known in the 19th century, if not earlier. I have no idea who first came up with the idea. $\endgroup$
    – Joe Silverman
    Commented Aug 19, 2021 at 23:29

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