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.
On
Yet an other answer following the same natural lines.
The cases when $x=0$ and/or $y=0$ are clear from the OP. We search for other, "new" solutions.
Let $a=(x,y)$ be the lcm of $x,y$, taken now to be $\ge 1$, so we can and do write $x=aX$, $y=aY$ with relatively prime integers $X,Y$. Then the given equation is equivalent to: $$ a(X+Y)(X^2-XY+Y^2) = X^2+42XY+Y^2\ . $$ This implies:
$(X+Y)$ divides $40XY=(X^2+42XY+Y^2)-(X+Y)^2$.
So $\color{brown}{(X+Y)}$ divides $\color{brown}{40}$.
$(X^2-XY+Y^2)$ divides $43XY=(X^2+42XY+Y^2)-(X^2-XY+Y^2)$.
So $\color{brown}{(X^2-XY+Y^2)}$ divides $\color{brown}{43}$.
(If a prime $p$ divides $(X+Y)$ and $XY$, then it divides either $X$ or $Y$, possibly exchanging notations let $p$ divide $X$, the $p$ divides also $(X+Y)-X=Y$, so $p$ divides the lcm $(X,Y)=1$, contradiction. The other case is similar.)
From here, some few cases have to be checked:
The first restriction lets us linearly substitute $Y$ in terms of $X$. The substitution $Y=d-X$, inserted in the second restriction gives $3X^2-3dX+d^2\in\{1,43\}$. thus getting some few equations of second degree in $X$ of the shape: $$ \begin{aligned} 3X^2 - 3dX + (d^2-1) &=0\ ,\\ 3X^2 - 3dX + (d^2-43) &=0\ . \end{aligned} $$ We obtain $(X,Y)$-solutions iff the discriminant is a square. So either $(4-d^2)$, or $(172-d^2)$ is a square.
Only the values $d=\pm 1$ and $d=\pm 2$ make the first expression ($1^2$ and $0^2$) a square.
Only the values $d=\pm 5$ and $d=\pm 8$ make the second expression ($7^2$ and $6^2$) a square.
We may and do assume $d>0$ (possibly changing now $a$ into $-a$), since passing from $d$ to $-d$ means replacing $(X,Y)$ by $(-X,-Y)$ (and $a$ by $-a$), which leads finally to the same final solution.
Here are these four cases explicitly displayed, and the four final solutions.
On
COMMENT.-For $x=y$ one has $2x^3=44x^2$ so $x=0$ and $x=22$. Besides we have $(x+y)^3-3xy(x+y)=(x+y)^2+40xy\iff S^3-3SP=S^2+40P$ where $S$ is sum and $P$ is product. Note that if $(x,y)$ is solution so is $(y,x)$.
We have $S|40P$, and $P=\dfrac{S^3-S^2}{3S+40}$. We consider first $S|40$ so $S=1,2,4,5,8,10,20,40$.
$$S=1\Rightarrow 43P=0\Rightarrow (x,y)=(1,0)\\S=2\Rightarrow4=46P\\S=4\Rightarrow P=\frac{48}{52}\\S=5\Rightarrow P=\frac{100}{55}\\S=8\Rightarrow P=\frac{448}{64}=7\Rightarrow X^2-8X+7=0\Rightarrow (x,y)=(7,1)$$ $S=10,20,40$ gives $P=\dfrac{900}{70},76,390$ respectively and the two last integers give $x=10\pm\sqrt{24}$ and $x=20\pm\sqrt{10}$.It remains to look at other possibilities
Thus we have found the solutions $(0,0),(22,22),(1,0),(7,1)$ given above by other users. The other solution is given taking $S=-5$ which gives $P=\dfrac{-150}{25}=-6$ from which the equation $X^2+5X-6=0$ so $(x,y)=(-6,1)$.
Here we want just to give another way of calculation.
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.