Prove: if $px^2 - (p+q)x + p = 0$ has an integer root, $px^2 + qx + p^2 - q = 0$ has an integer root too, for coprime $p,q \in \mathbb{N}$ For $\mathbf{coprime}$ $p,q \in \mathbb{N}$, how can I prove that:

if equation 

$$px^2 - (p+q)x + p = 0$$

has a root $x \in \mathbb{Z}$,

$$px^2 + qx + p^2 - q = 0$$ 

has an integer root as well?

Couldn't get a working idea.
In this answer, $p$ and $q$ are coprime integers such that $q\geq 0$ (and $p$ can be zero or negative).  We want to show that if $px^2-(p+q)x+p=0$ has an integer solution, then so does $px^2+qx+p^2-q=0$.

If $x=t$ is an integer root of $px^2-(p+q)x+p=0$, then
$$t(p+q-pt)=p.$$
If $t=0$, then $p=0$.  Because $\gcd(p,q)=1$, this means $q=1$.  Then, the equation $px^2+qx+p^2-q=0$, which is equivalent to $x-1=0$, has an integer solution $x=1$.

From now on, suppose that $t\neq 0$.  That is, $p=st$ for some integer $s$.  Hence,
$$p+q-pt=\frac{p}{t}=s.$$
Therefore,
$$q=s+p(t-1)=s+st(t-1)=s(t^2-t+1).$$
Therefore, $s$ divides $p$ and $q$, which implies $s=\pm1$ as $p$ and $q$ are coprime.  Since $q\geq 0$, we get $s=1$.

Therefore, $p=t$ and $q=t^2-t+1$.  Hence, $$px^2+qx+p^2-q=tx^2+(t^2-t+1)x+\big(t^2-(t^2-t+1)\big)$$ so that
$$px^2+qx+p^2-q=tx^2+(t^2-t+1)x+(t-1)=(tx+1)(x+t-1).$$
That is, $x=1-t$ is an integer solution to the quation $px^2+qx+p^2-q=0$.

The condition $q\geq 0$ is necessary.  Note that $p=-2$ and $q=-3$ satisfies the condition that $px^2-(p+q)x+p=0$ has an integer solution, but $px^2+qx+p^2-q=0$ does not have an integer solution.
From the first equation, the product of the roots is 1. So they are both 1 or both -1.        If they are both 1, substituting x=1 into the first equation gives q=p.                        Since p and q are co-prime, p=1,q=1 or p=-1,q=-1                                              If they are both -1, substituting x=-1 into the first equatio gives q=-3p.                  Since p and q are co-prime, p=1, q=-3 or p=-1, q=3                                           In all four cases for p and q, substituting the values for p and q into the second equation gives solutions for x that are integers.