<s>[INST] Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$ Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$
I try $x=0$ We have: $y^3-y^2=0 \Longrightarrow \left\{\begin{array}{l}
y=0 \\
y=1
\end{array}\right.$
I think, this equation only $(x,y)\in ${ $(0,0),(0,1),(1,0)\}$ but I can't prove that.
 [/INST] With the equation
$$x^3 + y^3 = x^2 + y^2 + 42xy \tag{1}\label{eq1A}$$
you've already handled the cases of $x = 0$ or $y = 0$. For the non-zero cases, have
$$\gcd(x,y) = d, \; x = de, \; y = df, \; \gcd(e, f) = 1 \tag{2}\label{eq2A}$$
Then \eqref{eq1A} becomes
$$\begin{equation}\begin{aligned}
(de)^3 + (df)^3 & = (de)^2 + (df)^2 + 42(de)(df) \\
d^3(e^3 + f^3) & = d^2(e^2 + f^2 + 42ef) \\
d(e + f)(e^2 - ef + f^2) & = e^2 + f^2 + 42ef
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This shows that $e^2 - ef + f^2$ must divide the right side, so using this and $e^2 + f^2 \equiv ef \pmod{e^2 - ef + f^2}$, gives
$$e^2 + f^2 + 42ef \equiv ef + 42ef \equiv 43ef \pmod{e^2 - ef + f^2} \tag{4}\label{eq4A}$$
Due to $\gcd(e,f) = 1 \implies \gcd(ef, e^2 - ef + f^2) = 1$, then
$$e^2 - ef + f^2 \mid 43 \tag{5}\label{eq5A}$$
Since $d$ is the $\gcd$, it'll be positive. If both $e$ and $f$ are negative, then there's no solution (since the left side of \eqref{eq1A} will be negative & the right side would be positive). Thus, either $e$ and $f$ are both positive or one is negative with the other being positive.
Either case gives $e^2 - ef + f^2 \gt 0$ so, from \eqref{eq5A}, this means $e^2 - ef + f^2 = 1$ or $e^2 - ef + f^2 = 43$. For it being equal to $1$, we must have $e$ and $f$ being positive and, since $e^2 - 2ef + f^2 = (e - f)^2 = 1 - ef \ge 0 \implies ef = 1$, then $e = f = 1$ is the only possible solution. This gives in \eqref{eq3A}
$$d(1 + 1)(1 - 1 + 1) = 1 + 1 + 42 \implies d = 22 \tag{6}\label{eq6A}$$
This shows $(x, y) = (22, 22)$ is a solution, as wimi's question comment states. Next, consider the other case of
$$e^2 - ef + f^2 = 43 \implies e^2 - (f)e + (f^2 - 43) = 0 \tag{7}\label{eq7A}$$
Treating $f$ as being constant, then solving for $e$ using the quadratic formula, gives a discriminant of $f^2 - 4(f^2 - 43) = 172 - 3f^2$, which must be $\ge 0$ so $\lvert f \rvert \le 7$, and is a perfect square. Trying the various integer possibilities shows that $f \in \{\pm 1, \pm 6, \pm 7\}$. Using these values to check $e = \frac{f \pm \sqrt{172 - 3f^2}}{2}$, then $d$ from \eqref{eq3A} and $(x,y)$ from \eqref{eq2A} (which I'll leave to you to do), results in the remaining non-zero solutions of $(x, y)$ are $\{(-6, 1)$, $(1, -6)$, $(1, 7)$, $(7, 1)\}$, as Christian Blatter's question comment indicates.
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