Integer solutions of an exponential equation

Question

How can I solve this equation?

$$7^{x} +2=y^{2}$$

$x$ and $y$ must be natural numbers.

Answer 1

A general method, not necessarily best for this particular equation, is to split into three cases by writing $x=3u+v$ with $v\in\{0,1,2\}$. Then rewrite your equation as $$ A(7^u)^3 + 2 = y^2\quad\text{with $A\in\{1,7,49\}$.} $$ So any solution to your equation gives an integer solution to one of the three equations $$ w^3+2=y^2,\quad 7w^3+2=y^2,\quad49w^3+2=y^2.$$ There are then standard methods for finding integer points on these genus $1$ curves.

Addendum: You can use the LMFDB to finish the solution. I'll rewrite the curves using the more typical $x,y$ variables.
For $y^2=x^3+2$, we see from http://www.lmfdb.org/EllipticCurve/Q/1728/n/4 that the curve has rank $1$ and the only integer solutions are $(-1,\pm1)$, which don't result in roots for your equation.
For $7w^3+2=y^2$, we multiply by $7^2$ and change coordinates to get the elliptic curve $y^2=x^3+98$. Then http://www.lmfdb.org/EllipticCurve/Q/28224/dp/2 says that the rank is $1$, and the only integer points are $(7,\pm21)$, which gives the solution $7^1+2=(\pm3)^2$.
And for $49w^3+2=y^2$, we multiply by $7^4$ and change coordinates to get the elliptic curve $y^2=x^3+4802$. Then http://www.lmfdb.org/EllipticCurve/Q/84672/fl/2 tells us that this curve has rank $0$ and no integer points. Hence your original equation $7^x+2=y^2$ has only the solutions $(x,y)=(1,\pm3)$.

Answer 2

Here's a proof that the $x=1$ solution is unique using only facts about "Pell equations" that were already known to Fermat (if not centuries earlier to Bhaskara II et al.) and should generalize at least to $p^x + 2 = y^2$ for odd primes $p$ such that $p+2$ is a square. I'll simplify the formulas by using $\sqrt 7$ (more generally, $\sqrt p$).

We start by observing that $x$ must be odd; this can be seen in various ways, for example by reducing $\bmod 4$. So, $7^x = 7 z^2$ where $z = 7^{(x-1)/2}$, and we seek solutions of $y^2 - 7 z^2 = 2$ where $z$ is a power of $7$. Now the general solution of $y^2 - 7z^2 = 2$ in natural numbers $y,z$ is $(y_n,z_n)$ such that $$ y_n + z_n \sqrt7 = \alpha u^n \quad (n \geq 0) $$ where $\alpha = 3+\sqrt 7$ and $u$ is the fundamental unit $\alpha / \bar\alpha = 8 + 3\sqrt 7$. (Note that this gives $y_n - z_n \sqrt 7 = \alpha u^{-(n+1)}$, so the solutions coming from $n < 0$ only repeat those with $n \geq 0$.) Now if $x > 1$ then $7 \mid z$, which happens if and only if $7 \mid 2n+1$. But then $z_n$ is a multiple of $z_3 = 7 \cdot 617$, and $617$ is not a multiple of $7$. Therefore $z_n$ is never a power of $7$ other than 1, QED.

(Alternatively, having found the condition $7 \mid 2n+1$ for $7 \mid z_n$, we could continue by asking when $7^2 \mid z_n$, $\ 7^3 \mid z_n$, etc., finding that $7^e | z_n$ if and only if $7^e | 2n+1$, and then $z_n \geq z_{(7^e-1)/2}$ which is much larger than $7^e$.)