Let
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
be a cubic polynomial given by
![P(x)=ax^3+bx^2+cx+d](/media/m/8/e/c/8ec99db6f320a045206115169dff7322.png)
, where
![a,b,c,d](/media/m/7/6/0/7605ede133e1f767d3890e0bfffb7b7f.png)
are integers and
![a\ne0](/media/m/4/b/0/4b00543273b3ba9d9943d7dc714cff3c.png)
. Suppose that
![xP(x)=yP(y)](/media/m/0/b/4/0b45903db6d3f52ecc6508781d4d00b6.png)
for infinitely many pairs
![x,y](/media/m/f/b/6/fb60533620f22cd699e5b58ce9a646a4.png)
of integers with
![x\ne y](/media/m/3/f/a/3fad6d61973825595977760dd0f94faf.png)
. Prove that the equation
![P(x)=0](/media/m/c/3/8/c38a1e79b07e0fb9a8e5ffed2f194eb6.png)
has an integer root.
%V0
Let $P$ be a cubic polynomial given by $P(x)=ax^3+bx^2+cx+d$, where $a,b,c,d$ are integers and $a\ne0$. Suppose that $xP(x)=yP(y)$ for infinitely many pairs $x,y$ of integers with $x\ne y$. Prove that the equation $P(x)=0$ has an integer root.