Solve $y^2 = x^3 + 16$

Question

I have been trying to solve the equation $y^2 = x^3 + 16$ over the integers, but failed to do so.

This is my incomplete solution:

$$(y - 4)(y + 4) = x^3$$

Case 1

$$\gcd(y - 4, y + 4) = 1$$ Let $m$ and $n$ be coprime integers. $$y - 4 = m^3$$ $$y + 4 = n^3$$ $$n > m$$ $$x^3 = m^3n^3$$

$$n^3 - m^3 = 8$$ $$(n - m)(n^2 + nm + m^2)$$

If $2$ divides $y - 4$, then $2$ also divides $y + 4$, so $y - 4$ and $y + 4$ are not divisible by $2$.

$$n ≡ m ≡ 1 \pmod 2 \implies n^2 + mn + n^2 ≡ 1 \pmod 2$$

In order for the $n^3 - m^3$ to be equal to $2$, $n^2 + mn + m^2 = 1 \implies n = 1, m = -1$, but then $n^3 - m^3 ≠ 8$, so $\gcd(y - 4, y + 4) ≠ 1$

Case 2

$$\gcd(y - 4, y + 4) > 1$$ $$d := \gcd(y - 4, y + 4)$$ $$a = {y - 4 \over d}$$ $$b = {y + 4 \over d}$$ $$\gcd(a, b) = 1$$ $$x^3 = d^2ab \implies d \space\vert\space x$$

I did not know what to do next. I would be grateful for a simple explanation.

For a complete solution see the notes by K. Conrad, Theorem $3.1.$ The only solutions in integers are $(0,\pm 4)$. The proof is elementary (e.g., no odd cubes differ by $8$). — Dietrich Burde, Apr 03 '23 at 15:40
@JackBlack FYI, using an Approach0 search, there's the AoPS threads $y^2=x^3+16$, Integers solutions, Number Theory, etc. Also, the same equation is asked about in Given integers $x,,y$ s.t. $x^2-16=y^3$, show that $x+4$ and $x-4$ are perfect cubes, but only about that property ... — John Omielan, Apr 03 '23 at 15:58
@JackBlack (cont.) mentioned in the subject line. There's also several similar Mordell equation questions, e.g., Tough Mordell equation $x^3+9=y^2$, A square number and a positive cubic number which differ by six, etc. — John Omielan, Apr 03 '23 at 16:01

MathStackexchangeIsNotSoBad · Accepted Answer · 2023-04-03T16:04:50.863

Originally by Wave particle (a person on aops)

Edit: It's the Keith Conrad's proof

Rewrite the equation as $x^3=(y-4)(y+4)$. We do casework based off the parity of $x,y$. It is easy to see they will both be the same parity.

Case 1: Both $x,y$ are odd.

This means both $y-4$ and $y+4$ are perfect cubes since $\gcd(y-4,y+4)=1$ by Euclidean algorithm. We can't have two positive odd cubes that differ by $8$ so there are no solutions in this case.

Case 2: Both $x,y$ are even.

It is easy to see that if $2|x$ then $4|y$ because otherwise $v_2(LHS)>v_2(RHS)$. Let $y=4y'$ in the original equation and we have $16y'^2=x^3+16$. If $4\nmid x$ then we have $v_2(LHS) > v_2(RHS)$ again. Thus, $4|x$. Letting $x=4x'$ in the new equation gives $y'^2=4x'^3+1$, implying that $y'$ is odd. Letting $y'=2k+1$, $4k^2+4k+1=4x'^3+1$ which becomes $k^2+k=x^3$. This implies $k, k+1$ are both cubes by Euclidean algorithm. It follows that $k=-1,0$ are the only solutions. This results in $x=0$ which implies $y=\pm4$. Thus, our solutions are $(x,y)=(0,4)$ and $(0,-4)$.

Oh really...then they must have taken help from Keith's proof — MathStackexchangeIsNotSoBad, Apr 03 '23 at 16:04

Solve $y^2 = x^3 + 16$

Case 1

Case 2

1 Answers1