Are there integer solutions for $y$ when $y^2 = x^3 +0x + 7$?

Question

Long story short I am curious if there is a way to find all the cube roots that will result into square root when I add the number $7$ to the cube root. Or with simple example:

$5^3$ is $125$ but when I add $7$ to $125,$ the result is $132$...which doesn't have a square root (not integer...). So if $5$ doesn't work, then what number once put to $3$rd $n$ will give number that once added by $7$ will give square root?

Here is the motivation behind my question: This one (Elliptic curve equation example...):

$y^2 = x^3 +0x + 7$

or better:

$y^2 = x^3 + 7$

Has a solution: $y = \sqrt{x^3 + 7}$

And if x = any integer number without remainder (1,2,3,4,5,6...), how can I find all the values of y where y is also integer?

In other words: the "Square root" of $x^3 + 7$ should be really a square root (e.q. $16,25,49...$) and not decimal?

I am looking for a simple solution, possibly with the help of modulo/primes since $7$ is a prime (and I'm also hoping to use this programically later in python).

Also if such integer exists at all it seems to be a very big number. I just tried to see if I can modulo $y$ to $2$ and it seems that no integer solutions can be found (well on wolframalpha...) even with $2$ as modulo.

Answer 1

Paste the following two lines to this website and press "Evaluate".

E = EllipticCurve([0, 7])
E.integral_points()

You'll get the result that there is no integral point on this curve (or better, for this integral model).

This also appears in GTM 106, "The Arithmetic of Elliptic Curves" by Silverman, page 266, Proposition 7.1.

You'll find the proof there.

Answer 2

Are there integer solutions to

$y^2 = x^3 + 7? \tag 1$

A challenging and engaging problem in the classical spirit of the school of Diophantus.

I don't have a solution for it.

Rather, one might say I have "half a solution".

We notice that if $(y, x)$ satisfies (1), then $y$ and $x$ are of opposite parity; that is,

$y \; \text{odd} \Longleftrightarrow y^2\; \text{odd} \Longleftrightarrow x^3 = y^2 - 7 \; \text{even} \Longleftrightarrow x \; \text{even}, \tag 2$

$y \; \text{even} \Longleftrightarrow y^2\; \text{even} \Longleftrightarrow x^3 = y^2 - 7 \; \text{odd} \Longleftrightarrow x \; \text{odd}. \tag 3$

We may resolve the first case (2) as follows:

with $y$ odd and $x$ even, there exist $m, n \in \Bbb Z$ such that

$y = 2m + 1, \tag 4$

and

$x = 2n; \tag 5$

then (1) becomes

$4m^2 + 4m + 1 = 8n^3 + 7, \tag 6$

whence

$4m^2 + 4m - 8n^3 = 6 \Longrightarrow 4 \mid 6, \tag 7$

impossible.

Thus there is no such pair $(y, x)$.

The "second half" (3) appears to be much more difficult; indeed if we try

$y = 2m, \; x = 2n + 1, \tag 8$

we find

$4m^2 = 8n^3 + 12n^2 + 6n + 8$ $\Longrightarrow 4m^2 - 8n^3 - 12n^2 - 6n = 8, \tag 9$

which admits no such simple resolution.

As I said, I only have "half a solution"; but at least those seeking further need only consider the case (3).