Solve $x^2+2^{17}=y^3$ over the integers

Question

I am asked to solve the equation $x^2+2^{17}=y^3$ over the integers. The technique I was taught for such equations uses algebraic numbers and a little algebraic number theory/ring theory depending on how you look at it. However, I got stuck at some point because I have never had such big numbers as $2^{17}$ in this type of equations and I really don't know what to do.
So, I began by rewriting the equation as $(x+256\sqrt{-2})(x-256\sqrt{-2})=y^3$. This is good because $\mathbb{Z}[\sqrt{-2}]$ is an Euclidean domain, so it is an UFD in particular.
Now the next step I was taught is to look at $d=\operatorname{gcd}(x+256\sqrt{-2}, x-256\sqrt{-2})$. It is easy to see that $d | 512\sqrt{-2}$. Now $2$ and $(\sqrt{-2})^2$ are associates (I will denote this by $2\sim (\sqrt{-2})^2$, I don't know if this is standard), so $d\sim (\sqrt{-2})^{19}$.
Thus, $d\sim (\sqrt{-2})^{j}$ for $j\in \{0, 1, 2, ..., 19\}$. I could even eliminate the associate sign and just write $d=\pm (\sqrt{-2})^j$, but it doesn't really make much of a difference.
Now I suppose that I am to see which $j$ work and which don't. The case $j=0$ is the easiest one, so I tried assuming that $j\ge 1$ to see what happens. Then $\sqrt{-2}|d|x\pm 256\sqrt{-2}|y^3$. This easily implies that $x$ and $y$ are both even, so $x=2x_1$ and $y=2y_1$ for some integers $x_1$ and $y_1$.
Substituting into the equation this gives $x_1^2+2^{15}=2y_1^3$. Thus, $x_1^2$ is even, so $x_1$ is even. This implies that $4|x_1^2$, so $2|y_1^3$. From here we get that $2|y_1$, so $8|y_1^3$. After all of this, $16|x_1^2$, so $4|x_1$. Now we write $x_1=4x_2$ and $y_1=2y_2$ for some integers and we get that $$x_2^2+2^{11}=y_2^3.$$ Here is where I am stuck. This equation looks just like the one at the beginning. I could write $d=\operatorname{gcd}(8(x+32\sqrt{-2}), 8(x-32\sqrt{-2}))\sim 8\operatorname{gcd}(x+32\sqrt{-2}, x-32\sqrt{-2})$ and this basically tells me that if $j\ge 1$ then $j\ge 6$ in fact, but that's all I can do.

Answer 1

Note, your approach is about powers of $2$ that divide $x.$ How many factors of $\sqrt{-2}$ divide $x+512\sqrt{-2}$ is entirely about powers of two that divide $x.$

---

Review

Let $\nu_2(m)$ be the largest integer $n$ such that $2^n\mid m.$

A few key results:

$$\nu_2(m^k)=\nu_2(m)^k$$

and,

If $\nu_2(m_1)\ne \nu_2(m_2)$ then: $$\nu_2(m_1+m_2)=\min(\nu_2(m_1),\nu_2(m_2))$$

---

In your equation, $\nu_2(y^3)$ is a multiple of $3$ and, since $\nu_2(x^2)$ is even, it can't equal $\nu_2(2^{17})=17.$ So $$\nu_2(x^2+2^{17})=\min(2\nu_2(x),17).$$

So $\nu_2(x)\leq 8$ and must be divisible by $3.$

Likewise, $\nu_2(y)$ must be even.

So either:

1. $x$ is odd
2. $x=8x_1,$ $x_1$ odd.
3. $x=64x_2,$ $x_2$ odd.

So either:

1. $x$ is odd and $x+512\sqrt{-2}$ is a perfect cube in $\mathbb Z[\sqrt{-2}].$
2. $x_1$ is odd and $x_1+32\sqrt{-2}$ is a perfect cube
3. $x_2$ is odd and $x_2+4\sqrt{-2}$ is a perfect cube.

Solve each of these independently.

---

Expanding $(a+b\sqrt{-2})^3,$ you get the equation:

$$b(3a^2-2b^2)=2^{3k+2}$$ for some $a,b$ and $k=0,1,2.$

So $b=\pm 2^i$ for some $i\leq 3k+2.$ Plugging in and re-arranging, you get:

$$3a^2=2^{2i+1}\pm 2^{3k+2-i}$$ Since $\nu_2(3a^2)$ is even, and $3k+2-i\neq 2i+1$ you must have $2i+1>3k+2-i$ or $i>k,$ and $3k+2-i$ must be even. Modulo $3,$ $2^{2i+1}\equiv 2$ and $2^{3k+2-i}\equiv 1,$ so the sign in the expression is $+.$

So: $$3a^2=2^{3k+2-i}\left(2^{3(i-k)+1}+1\right)$$

And you need $k<i\leq 3k+2$ and $k\equiv i\pmod 2$ and $$2^{3(i-k)+1}+1=3a_0^2.$$

If $i=k+2j,$ this becomes $0<i\leq k+1$ and: $$2^{6j+1}+1=3a_0^2.$$

But none of $2^{7}+1, 2^{13}+1, 2^{19}+1$ is three times a square. In fact, all three are of the form $3p,$ where $p$ is prime.

So there are no solutions.