TITLE: If $x+y^3,x^2+y^2,x^3+y$ are all integers, are $x,y$ both integers?
QUESTION [11 upvotes]: Let $x,y$ be both real numbers. If $x+y^3,x^2+y^2,x^3+y$ are integers, are $x,y$ both integers?

This question begins with two real numbers while usual number theory tricks are based on the precondition that the variables are integers. Showing $x,y$ are algebraic numbers is easy by observing $x+((x^3+y)−x^3)^3$ is an integer, but how can algebraic numbers help?

REPLY [3 votes]: In this solution $x$ and $y$ are not assumed to be real numbers.  I will find all $(x,y)\in\Bbb C^2\setminus\Bbb Z^2$ such that $x^3+y$, $x+y^3$, and $x^2+y^2$ are integers.  The claim is that all possible $(x,y)$ are given below by $(3)$ with $|d|\ge2$, and $(4)$ for any integer $d$, and none of such solutions satisfy $(x,y)\in\mathbb{R}^2$.
Let $a=x^3+y$, $b=x+y^3$, and $c=x^2+y^2$.  Observe that $x$ is a root of
$$A(t)=(t^3-a)^3-t+b$$
and $y$ is a root of 
$$B(t)=(t^3-b)^3-t+a.$$
Therefore both $x$ and $y$ are algebraic integers.  
If $x=0$, or $y=0$, then it is easy to see that $x$ and $y$ are integers.  If $x= \pm y$, then $x^2=\frac{c}{2}$, so
$$a=x^3+y=\frac{c}{2}x\pm x=\frac{c\pm 2}{2}x.$$
If $c=2$, then $x^2=\frac{c}{2}=1$, so $x$ is an integer, making $y$ also an integer.  If $c=-2$, then $x^2=\frac{c}{2}=-1$ so $x=\pm i$.  Therefore we have $2$ solutions
$$(x,y)=\pm(i, i).\tag{1}$$
If $c\ne \pm2$, then $x$ is rational.  A rational algebraic integer is necessarily an integer.  Thus, $x$ and $y$ are both integers.
Suppose now that $x\ne \pm y$, $x\ne 0$, and $y\ne 0$. 
 If $a=b$, then
$$(x-y)(x^2+xy+y^2-1)=(x^3+y)-(x+y^3)=a-b=0.$$
Since $x\ne \pm y$, we get $x^2+xy+y^2=1$, so $$xy=1-x^2-y^2=1-c.$$
Hence $x^2$ and $y^2$ are roots of the quadratic
$$Q(t)=t^2-cx+(1-c)^2.$$
Note that $$x^3+y=x(x^2+y^2)-xy^2+y=cx-(1-c)y+y.$$
So
$$a=c(x+y).$$
If $c=0$, then $a=0$ so $b=0$.  Now, $Q(t)=t^2+1$, so $x^2,y^2=\pm i$.  Therefore
$$x,y=\pm\frac{1}{\sqrt2}\pm\frac{i}{\sqrt2}.$$
Because $x^2+y^2=c=0$, we get $4$ solutions:
$$(x,y)=\left(e^{i\frac{k\pi}{4}},e^{-i\frac{k\pi}{4}}\right),\tag{2}$$
where $k=\pm1,\pm3$.  If $c\ne 0$, then $x+y=\frac{a}{c}$ and $xy=1-c$.  Hence the algebraic integers $x$ and $y$ are roots of 
$$P(t)=t^2-\frac{a}{c}t+(1-c).$$
Note that this means $\frac{a}{c}$ is an integer.  We write $a=cd$.  Then
$$c=x^2+y^2=(x+y)^2-2xy=d^2-2(1-c)$$
so $$d^2=c+2(1-c)=2-c.$$
Hence
$$P(t)=t^2-dt+(1-c)=t^2-dt+(d^2-1).$$
The roots are
$$x,y=\frac{d\pm\sqrt{4-3d^2}}{2}.$$
Thus there are $2$ solutions
$$(x,y)=\left(\frac{d\pm\sqrt{4-3d^2}}{2},\frac{d\mp\sqrt{4-3d^2}}{2}\right).\tag{3}$$
Note that if $x$ and $y$ are both real, then $d=0$ or $d=\pm1$, but this means $x$ and $y$ are integers.  Non-integer solution given by $(3)$ must come from $|d|\ge 2$.  Solutions $(2)$ can be included in $(3)$ by taking $d=\pm 2$.
If $a=-b$, then
$$(x+y)(x^2-xy+y^2+1)=(x^3+y)+(x+y^3)=a+b=0.$$
Because $x\ne \pm y$, we get $x^2-xy+y^2=-1$ so
$$xy=x^2+y^2+1=c+1.$$
Hence $x^2$ and $y^2$ are roots of the quadratic
$$Q(t)=t^2-cx+(c+1)^2.$$
Note that $$x^3+y=x(x^2+y^2)-xy^2+y=cx-(c+1)y+y.$$
So
$$a=c(x-y).$$
If $c=0$, then $a=0$ so $b=0$, and we are back to solutions $(2)$. If $c\ne 0$, then $x-y=\frac{a}{c}$ and $x(-y)=-(c+1)$.  Hence the algebraic integers $x$ and $-y$ are roots of 
$$P(t)=t^2-\frac{a}{c}t-(c+1).$$
Note that this means $\frac{a}{c}$ is an integer.  We write $a=cd$.  Then
$$c=x^2+y^2=(x-y)^2+2xy=d^2+2(c+1)$$
so $$d^2=c-2(c+1)=-2-c.$$
Hence
$$P(t)=t^2-dt-(c+1)=t^2-dt+(d^2+1).$$
The roots are
$$x,-y=\frac{d\pm\sqrt{-4-3d^2}}{2}.$$
Thus there are $2$ solutions
$$(x,y)=\left(\frac{d\pm\sqrt{-4-3d^2}}{2},-\frac{d\mp\sqrt{-4-3d^2}}{2}\right).\tag{4}$$
None of the solutions given by $(4)$ are real.  Solutions $(1)$ can be included in $(4)$ by taking $d=0$.
Let now suppose that $x\ne \pm y$, $x\ne 0$, $y\ne 0$, and $a\ne \pm b$. 
$$a=x^3+y=(x^2+y^2)x+(1-xy)y=cx+(1-p)y$$
and
$$b=x+y^3=(1-xy)x+(x^2+y^2)y=(1-p)x+cy.$$
Since $a\ne \pm b$, $p\ne 1\pm c$.  If $p$ is an integer and $p\ne 1\pm c$, then the system above yields rational solutions $(x,y)$, which means that $x,y$ are integers.  We aim to show that $p$ is an integer.
If $c=0$, then $x^2+y^2=0$ so $y=\pm xi$.  We have
$$a=x^3+y=x^3\pm xi$$
and
$$b=x+y^3=x\mp x^3i=\mp (x^3\pm xi)i=\mp ai.$$
This can happen only when $a=b=0$, but this brings us back to solutions $(1)$.  We now suppose that $c\ne0$.
The rest uses the other answer by Carl Schildkraut.  If $p=xy$, then we have that $p=xy$ is a root of the polynomials
$$S(t)=t(t-1)^2-k$$
and
$$T(t)=t^3+t^2-t-r$$
for some $k\in\Bbb Z$ and $r\in\Bbb Q$.  Thus $p$ is a root of
$$M(t)=\frac{T(t)-S(t)}{3}= t^2-\frac23t+\frac{k-r}{3}.$$
Note that $p$ is an algebraic integer because it is a root of $S(t)$.  Therefore $p$ is a root of a monic irreducible polynomial $R(t)\in \Bbb Z[t]$.  We must have $R(t)\mid M(t)$.  Since $M(t)\notin\Bbb{Z}[t]$, $R(t)\ne M(t)$, so $R(t)$ must be linear.  That is, $p$ is an integer.