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I am aware of Legendre's method for solving the Quadratic Diophantine Equation (QDE) of the form

$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \tag{1} $$

by transforming them into a Pell-type equation.

I have the following equation:

$$ x^3+4xy-8(n+1)y+4(n+1)^2=0 \tag{2} $$

where $n$ is a constant.

The leading monomial is of degree $3$. So it isn't a QDE.

Are there any methods for solving equations of type (2) with a cubic term?

vvg
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    This particular one is linear on $y$. Solve for $y$, which should be equal to a rational functions of $x$. The denominator is linear too. So, dividing numerator by denominator would give a polynomial plus a proper fraction with a constant on the numerator. There is the minor inconvenience of non-monic denominator, but appropriate multiplication by $4$ and study of cases when that multiplication introduces extra solutions should make it work. – NDB Aug 02 '23 at 21:07
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    $y=\frac{x^3+4(n+1)^2}{-4x+8(n+1)}$. We need $x$ even and $n$ odd. So, let's change $x$ with $2z$ and $n$ with $2k+1$. We get $y=\frac{z^3+2(k+1)^2}{-z+2(k+1)}$. Well, now the denominator is monic, easier to do the division. – NDB Aug 02 '23 at 21:11
  • @NDB: If you could write this up as an answer, I will accept it and move it off the unanswered queue. Thanks. – vvg Aug 04 '23 at 03:59
  • @vvg What is the context of this post? It has a nice expression in terms of powers, $$x^3 + x^2 - (x - 2 y)^2 + 4 (1 + n - y)^2 = 0$$ so doesn't seem to be a random equation. – Tito Piezas III Nov 20 '23 at 16:39

1 Answers1

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The given equation is linear in $y$, so we can readily isolate it to get $$y=\frac{x^3+4(n+1)^2}{8(n+1)-4x}.$$ For $y$ to be an integer $x$ and $n+1$ must be even, say $x=2z$ and $n=2m-1$, and so $$y=\frac{z^3+2m^2}{2m-z}=\frac{2m^2(4m+1)}{2m-z}-(4m^2+2mz+z^2).$$ Given that $n$ is a constant, also $2m^2(4m+1)$ is a constant, and now it suffices to check all values of $z$ such that $2m-z$ divides $2m^2(4m+1)$.

Servaes
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