What would be the cardinality of $S$? Let $S$ be the set of polynomials $f(x)$ with integer coefficients satisfying $$f(x)\equiv 1\mod (x-1)$$ and $$f(x)\equiv 0\mod (x-3)$$
What would be the cardinality of $S$?
Second equation means $x=3$ is a root of $f(x)$. That means $f(3)\equiv 1\mod2 $
 and that means $0\equiv 1\mod 2$ which is a contradiction. So $S$ is empty.
Please have a look at the work. Am I right?
Thanks.

Ok, after the very last comment below the question by the OP I think the following can be what was intended: it is given $\;f(x)\in\Bbb Z[x]\;$ s.t. both the following are true:
$$\begin{cases}I\;\;\;f(x)\cong1\mod{(x-1)}\iff f(x)=1+(x-1)p(x)\;,\;\;p(x)\in\Bbb Z[x]\\{}\\
II\;\;f(x)\cong0\mod{(x-3)}\iff f(x)=(x-3)q(x)\;,\;\;q(x)\in\Bbb Z[x]\end{cases}$$
but then we get:
$$\begin{cases}I\;\implies\; f(1)=1=1\pmod 2\\{}\\
II\implies f(1)=-2q(1)=0\pmod2\end{cases}\;\;\;\implies\text{ contradiction}$$
and what the OP did is thus correct: we get that in fact $\;S=\emptyset\;$ .
