<s>[INST] If $x^3+y^2$ and $x^2+y^3$ are integers, show that $x$, $y$ are integers 
Given rational numbers $x$, $y$ such that $x^3+y^2$ and $x^2+y^3$ are integers, prove that $x$, $y$ are integers.

For this problem I don't even know how to start tackling it. I tried various ways:

*

*Letting $x = \frac{a}{b}$, $y = \frac{c}{d}$, which only makes it more complicated

*Doing operations on them: $x^3+y^2+x^2+y^3=x^2(x+1)+y^2(y+1)$, which I have no idea how to continue;

$(x^3+y^2)(x^2+y^3)=x^5+y^5+x^3y^3+x^2y^2$, and it's also probably too complex to break down
$x^3+y^2-x^2-y^3=(x-y)(x^2+xy+y^2)+(y-x)(x+y)$ seems more plausible, but I also can't do anything with this.
I would really appreciate any way of tackling this problem, because I have spent a while thinking about this problem.
EDIT: I have found a solution to this problem (this was edited 2 days after this was posted, so there were also answers before this):

Suppose $x = \frac{a}{b}$, $y = \frac{c}{d}$ in which $a$, $b$, $c$, $d$
are integers and $\gcd(a,b) = \gcd(c,d) = 1$. Then we have
$x^2+y^3=\frac{a^2}{b^2}+\frac{c^3}{d^3}=\frac{a^2d^3+b^2c^3}{b^2d^3}$,
so:
$b^2|a^2d^3+b^2c^3$, which means $b^2|a^2d^3$, and $\gcd(a,b)=1$ so
$b^2|d^3$, and
$d^3|a^2d^3+b^2c^3$, which means $d^3|b^2$, so $b^2=d^3$.
Doing the same to $x^3+y^2$, we get $b^3=d^2$. From that we get
$b^5=d^5$, so $b=d$. Substituting to $b^2=d^3$, we get $b^2=b^3$,
hence $b=d=1$, which implies that $x$ and $y$ are integers.

 [/INST] Suppose $x=a/d$ and $y=b/d$ with $d\gt0$ and $\gcd(a,b,d)=1$ (i.e., write $x$ and $y$ with the smallest possible common denominator).  Then $x^2+y^3=m$ implies $b^3=(md^2-a^2)d$, which implies $d\mid b^3$. Likewise, $x^3+y^2=n$ implies $d\mid a^3$. Thus
$$0\lt d=\gcd(a^3,b^3,d)\le\gcd(a^3,b^3,d^3)=(\gcd(a,b,d))^3=1^3=1$$
hence $d=1$, and so $x$ and $y$ are integers.
</s>