Can $p(x)q(x+1)$ and $q(x)p(x+1)$ be the same? Suppose $p(x)$ and $q(x)$ are distinct polynomials with integer coefficients. Can we conclude that $p(x)q(x+1)$ and $q(x)p(x+1)$ are distinct expressions?
 A: No, you can't conclude that they're necessarily distinct expressions. An extremely simple example is if $p(x) = a$ and $q(x) = b$ where $a$ and $b$ are different integer constants, so then $p(x)q(x + 1) = q(x)p(x + 1) = ab$. A non-constant polynomial example is to let $p(x) = x$ and $q(x) = -x$. Then $p(x)q(x + 1) = x(-(x + 1)) = -x(x + 1)$ and $q(x)p(x + 1) = -x(x + 1)$.
As Somos suggested in a question comment, consider that $q(x) \neq 0$ so can use the rational function $f(x) = \frac{p(x)}{q(x)}$. Then, dividing both sides of $p(x)q(x + 1) = q(x)p(x + 1)$ by $q(x)q(x + 1)$ gives $\frac{p(x)}{q(x)} = \frac{p(x+1)}{q(x+1)}$, i.e., $f(x) = f(x + 1)$. For this to be true for all $x$ requires that $f(x)$ be a constant function (note that, as Jyrki Lahtonen wrote in a question comment, he gives a proof of this in his answer at Intersection of two subfields of the Rational Function Field in characteristic $0$), i.e., $f(x) = c$ so $p(x) = cq(x)$ for some rational constant $c \neq 1$ such that all of the coefficients in $p(x)$ are integers. In my $2$ examples above, $c = \dfrac{a}{b}$ (with $b \neq 0$) and $c = -1$. In the special case of $q(x) = 0$, then we have $q(x) = cp(x)$ instead, with $c = 0$.
A: We can prove the following theorem which essentially answers your question:

If $p(x)$ and $q(x)$ are distinct monic polynomials with complex coefficients and
  $$p(x)q(x+1)=q(x)p(x+1)$$
  then $p=q$.

This has an easy converse too:

If $p(x)$ and $q(x)$ have either that $p=cq$ or $q=cp$ for some complex $c$, then
  $$p(x)q(x+1)=q(x)p(x+1)$$

Together, these classify entirely the solutions to your equation.
To prove the offered theorem, we first note that if a pair $p(x)$ and $q(x)$ have this property and $r(x)$ is any polynomial dividing both, then $\frac{p(x)}{r(x)}$ and $\frac{q(x)}{r(x)}$ also must have the given property, as both sides are divided by $r(x)r(x+1)$. Thus, if there is a counterexample, there is a counterexample where $p(x)$ and $q(x)$ are coprime. We assume now that $p(x)$ and $q(x)$ are coprime, so share no roots.
Let $r$ be a root of $p(x)q(x+1)=q(x)p(x+1)$ with maximal real part. Note that $r$ cannot be a root of $q(x+1)$, as then $r+1$ is a root of $q(x)$. Similarly, $r$ is not a root of $p(x+1)$. Thus, $r$ must be a root of both $p(x)$ and $q(x)$. This contradicts coprimality. Thus, to the contrary, $p(x)q(x+1)=q(x)p(x+1)$ must have no roots so $p(x)=q(x)=1$.
A: (1). Suppose $p,q$ are real polynomials with $deg(p)\ge deg(q)>0$ and $\forall x\in \Bbb R\,(\,p(x)q(x+1)=p(x+1)q(x)\,).$ 
We have $p=qr+s$ where $r,s$ are polynomials with $deg(s)<deg(q).$
Let $A=\{x\in \Bbb R: \forall x\in \Bbb N \,q(x+n)\ne 0\}.$ Observe that $A$ is an infinite set (since the polynomial $q$ is not the constant $0\,$) a fact to be used later.  
For $x\in A$ let $B(x)=\frac {p(x)}{q(x)}.$ Then for $x\in A$ we have $B(x)=B(x+n)$ for all $n\in \Bbb N,$ so $$B(x)=\lim_{n\to \infty;n\in \Bbb N} B(x+n)=$$ $$=\lim_{n\to \infty;n\in \Bbb N} r(x+n) +\frac {s(x+n)}{q(x+n)}.$$ Now $\lim_{n\to \infty}\frac {s(x+n)}{q(x+n)}=0$ because $deg (q)>deg (s).$ Also $\lim_{n\to \infty}r(x+n)$ cannot exist for the polynomial $r$ unless $r$ is a constant. So let $r(y)=K\in \Bbb R$ for all  $y\in \Bbb R.$ Now we have $$\forall x\in A\;( B(x)=K).$$ Therefore $$\forall x\in A\,(p(x)-Kq(x)=0).$$ So the polynomial $p-Kq$  is $0$ on the infinite set $A.$ Therefore $$\forall x\in \Bbb R\;(p(x)=Kq(x)).$$
Now observe that the converse $(p=Kq \land deg(p)\ge deg(q)>0)$ implies $\forall x\in \Bbb R \,(p(x)q(x+1)=p(x+1)q(x)\,).$
Regarding constant polynomials: If $q=0$  then (obviously) $p(x)q(x+1)=p(x+1)q(x).$ If $q $ is a non-$0$ constant then $\forall x \in \Bbb R(\,p(x)q(x+1)=p(x+1)q(x)\,)\implies \forall x\in \Bbb R\,(p(x+1)=p(x)\,)$  which implies the polynomial $p$ is constant. 
