$x^2 + y^3 = 2024$ over the integers

Question

How can you find all solutions to the equation $x^2 + y^3 = 2024$ over the integers?

This question came to me randomly, and checking with technology, there are a total of ten solutions:

$x = \pm 444, y = -58$

$x = \pm 140; y = -26$

$x = \pm 45; y = -1$

$x = \pm 41; y = 7$

$x = \pm 32; y = 10$

Is there a better way to go about this rather than bashing? I see you could try to factor the equation by adding one and then making a difference of squares like $$y^3+1=45^2-x^2,$$ but I don’t know how to proceed from there.

FYI, using an Approach0 search, there's the related Integer solutions for $x^2-y^3 = 23$. (although in this case, there are no integer solutions, and the answer shows that), and Solution to $x^3-y^2=6$. There's also the AoPS thread Number theory. — John Omielan, Oct 16 '24 at 04:24
Maybe start by thinking of the integer partitions of 2024 of length 2 (I.e. $(2023,1), (2022,2), ...(2024-k,k)$ ? And finding which $k$ are perfect square/cubes? — ydd, Oct 16 '24 at 04:25
The presence of $2024$ suggests that this is a challenge question and maybe a current one. — badjohn, Oct 16 '24 at 04:30
Try the following expression: $n^3+T_{n-1}^2=T_{n}^2$ with $T_{n} a triangular number. — user25406, Oct 16 '24 at 11:30
$(y^3 + 1) = (y+1)(y^2 -y +1) = 45^2 -x^2 = (45-x)(45+x)$. My above comment on using triangular numbers wouldn't work because $2024$ is not a square. — user25406, Oct 16 '24 at 13:16
@user25406: I agree with your factorizations, but how do they help? — J. W. Tanner, Oct 16 '24 at 16:35
@J.W.Tanner, It's worth trying the two possibilities. 1- $y+1=45-x$ and use the $y=44-x$ in the original equation and solve the cubic equation in $x$. 2- $y+1=45 + x$ and do the same thing. I am not sure it will get us a solutions. — user25406, Oct 16 '24 at 17:51
@user25406: also just because $ab=cd$ doesn’t mean $a=c$ or $a=d$ — J. W. Tanner, Oct 16 '24 at 18:18
Graphing $y^3 +1$ and $45^2-x^2$ provide only one non-integer solution for the $y^3 +1= 45^2-x^2$ which is not exactly the original equation. — user25406, Oct 16 '24 at 20:27
The related equation $y^2 = x^3 + n$, which is the form of a seemingly more extensively studied equation (called a Mordell Curve) is very similar to this. It is known there are finitely-many solutions to these equations, and the precise solutions are known for a large range of values of n (up to $10^7$, at least). For $n = 2024$, the solutions are practically identical to this equation (with a swap of variables). — Avery Wenger, Oct 17 '24 at 03:38
Cf. Mordell curve on wikipedia and this page under E_+02024. — J. W. Tanner, Oct 21 '24 at 01:10

score -1 · Answer 1 · answered Oct 28 '24 at 15:14

Comment: Suppose the author of the question expect an elementary number Theoric solution at high school level, then we may argue:

Case 1: $(x, y, 3)=1$, then we have:

$x^2\equiv 1\bmod 3$

$y^3\equiv y\bmod 3$

$\Rightarrow x^2+y^3\equiv (1+y)\bmod 3$

$2024\equiv 2 \bmod 3$

$\Rightarrow y+1\equiv 2\bmod 3$

$y\equiv 1 \bmod 3$

Now we may write:

$(y, y^3)=(4, 64), (7, 343), (10, 1000), (13, 2197)$

So only numbers$ 4$ , $7$ and $10$ can be checked, we can see that only numbers 7 and 10 give integer for x:

$y=7\rightarrow x=41$

$y=10\rightarrow x=32$

Also:

If $a\equiv -1\bmod 3\Rightarrow -a\equiv 1 \bmod 3$

$(y, y^3, -1\bmod 3)=(2, 8, 7), (3, 27, 26), (7, 343, 342)\cdot\cdot\cdot$

For example $-26$ gives $x=140$

There are probably more solutions with negative $y$.

Case 2:

$x\equiv 0\bmod 3$

$y^3\equiv y\bmod 3$

$x^2+y^3\equiv (0+y)\bmod 3\equiv 2\bmod 3$

$\Rightarrow y\equiv 2\bmod 3\equiv -1\bmod 3$

We may write:

$(y, y^3)=(-1, -1), (-4, -64), (-7, -343), \cdot\cdot\cdot (-58, -195112)\cdot\cdot\cdot$

$-1$ gives $x=45$ and $-58$ give $444$.

There are probably more solutions with negative $y$.

$x^2 + y^3 = 2024$ over the integers

1 Answers1