Solving a non-linear Diophantine equation I'm having some problems solving this Diophantine equation so I'd really appreciate any small hints to guide me along:
Solve for $(x,y)$ over the integers in $y(x+y)=x^{3}-7x^{2}+11x-3$.
I'm first trying to get something out of the quadratic formula so I write:
$y^{2}+xy-x^{3}+7x^{2}-11x+3=0$
$\frac{-x\pm\sqrt{x^{2}+4x^{3}-28x^{2}+44x-12}}{2}$ is an integer, so eventually $4x^{3}-27x^{2}+44x-12=A^{2}$, where $A$ is some integer, so $(A-x)(A+x)=x^{3}-7x^{2}+11x-3=y(x+y)$. I write this as another quadratic equation so $x^{2}+xy+y^{2}-A^{2}=0$ and $\frac{-y\pm\sqrt{y^{2}-4y^{2}+4A^{2}}}{2}$ is an integer. So $3y^{2}=4A^{2}-B^{2}$, where $B$ is some integer, well I do this another time for x and I get $3x^{2}=4A^{2}-C^{2}$, where $C$ is another integer. And this is where I'm not sure what to do.
Would be great if someone just gave me a small clue (please don't post the whole solution yet!). Thanks.
 A: Here’s one way you might move forward with your approach.
You found that
\begin{align} \tag{$\star$}
4x^3-27x^2+44x-12 &= (x-2)(4x^2-19x+6)
\end{align}
must be a square. Note that
$$4x^2-19x+6 = (4x-11)(x-2)-16.$$
This can be handled in two cases: $x$ odd [in which case the absolute value of each factor on the right-hand side of $(\star)$ must be the square of an integer], and $x$ even [where the two factors have a common factor].
I hope that’s enough of a hint without solving the problem for you!
A: Your equation $\,y(x+y)=x^{3}-7x^{2}+11x-3\,$ is an elliptic
curve in Weierstrass form. Let
$$f(x,y) := -y(x+y)-7x^{2}+11x-3.$$ Then
$$f(x+2,y-1) = -y(x+y) + x^3 -x^2 -4x.$$
This is the equation of the Elliptic curve with LMFDB
label 530.b1
whose five integer points are given there.
I don't think that elementary methods such as you
propose will be able to do what you want.
A: You know that $4x^3-27x^2+44x-12$ is a square. Let $x=u+2$ and then $(4u^2-3u-16)u$ is a square.
Now let $u=tR^2$ where $R$ is a positive integer and $t$ is square-free. Then $4u^2-3u-16=tS^2$ where $S$ is a positive integer and $4tR^4-3R^2-16/t=S^2$. Note that the only possibilities for $t$ are $-2,-1,1,2$.
HINT for -ve cases.   We have $S^2\le16$. You should obtain the solutions $(1,-2),(1,1),(2,-1)$.
HINT for $t=2$. Remember that squares are $0,1$ or $4$ modulo $8$. Prove that $R$ and $S$ are both even and then repeat the argument to obtain a contradiction.
HINT for $t=1$. We have $4R^4-3R^2-16=S^2$. Completing the square gives $$(8R^2-3)^2-(4S)^2=265.$$ Now express $265$ as a difference of squares (there are two possibilities). You will find that one of the possibilities leads to the solutions $(6,-3),(6,-9)$.
