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In studying Diophantine equations, the following was thrown up (as an example):

$my^2 = 28x^3 - 5x^2 + 2805x + 2323$

I am trying to solve for $x$ where $x$, $m$ and $y$ are all unknown.

Another similar example is:

$my^2 = 76x^3 - 1639x^2 + 515618x + 224202$

In the first example $x = 3$ or $26$ are both solutions,

and in the second example $x = 58$ or $116$ are both solutions.

The underlying structure conforms to the following equation:

$my^2 = 4px^3 +(n-4pq)x^2 + (4pr-qn)x + rn$,

wherein: $p$ is a prime number, such that if $y_1$.$y_2$ = $N$, and $N$ is the number to be factorised, then $y_1$ and $y_2$ are its factors.

$n$ is calculated as the modular square root of $N$, then $q$ and $r$ are found using the Euclidean algorithm as follows:

$(N-n^2)/4p = (4p)r + (n)q$

In the first example, $N = 80357$; and since $p = 7$ is a factor of $(y_2 - y_1)/2$, then $n$ is determined to be $23$ via Tonelli-Shanks's modular square root algorithm. We now find that $(N-n^2)/4p = (4p)r + (n)q$, or:

$(80357-23^2)/(28) = (28)r + (23)q$

i.e. $2851 = (28)r + (23)q$

is satsified by $(r,q) = (101,1)$

Hence, after algebraic manipulation the first equation is generated. The second follows the same process, but here $N = 39298409$. I don't have any experience in solving elliptic curve -type equations; what would be the best way to proceed from here?

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    $\begingroup$ What's your question, exactly? Clearly every $x\in\mathbb Z$ yields a solution, e.g. take $y=\pm1$ and calculate $m$. $\endgroup$ Commented Dec 1, 2024 at 18:33

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