$f(x) \equiv 1$ mod $(x-1)$ and $f(x) \equiv 0$ mod $(x-3)$ then is there any $f(x)$? Let $S$ be the set of polynomials $f(x)$ with integer coefficients satisfying 
$f(x) \equiv 1$ mod $(x-1)$
$f(x) \equiv 0$ mod $(x-3)$
Which of the following statements are true?
a) $S$ is empty .
b) $S$ is  a singleton.
c)$S$ is  a finite non-empty set.
d) $S$ is  countably infinite.
My Try: I took $x =5$ then $f(5) \equiv 1$ mod $4$ and $f(5) \equiv 0$ mod $2$ . Which is impossible so $S$ is empty.
Am I correct? Is there any formal way to solve this?
 A: Let we have a general polynomial form as following$$f(x)=a_0+a_1x+\cdots +a_dx^d$$We know that for some polynomials $a(x)$ and $b(x)$ $$f(x)=1+(x-1)a(x)=b(x)(x-3)$$therefore $$f(1)=1\\f(3)=0$$ which leads to $$(1)\quad a_0+a_1+\cdots +a_d=1\\(2)\quad a_0+3a_1+\cdots +3^da_d=0$$by subtracting $(1)$ from $(2)$ we obtain $$2a_1+8a_2+\cdots +(3^d-1)a_d=-1$$which is impossible since the LHS is even and the RHS is odd, so such a polynomial doesn't exist
A: The second condition basically says that $3$ is a root. 
But then substitute $x=3$ into the first condition: $0=f(3)\equiv 1 \pmod 2$, a contradiction.
A: Since $f(x)\equiv f(n)\mod x-n$, the second condition says $f(3)=0$ and $f(x)=(x-3)g(x)$ for some $g(x)\in\mathbf Z[x]$. But then $\;f(1)=-2g(1)$ is even, contrary to the first condition $f(1)=1$.
A: We have that
$$f(x)\equiv 1\bmod{(x-1)}\iff \exists a(x)|f(x)=(x-1)a(x)+1.$$
$$f(x)\equiv 0\bmod{(x-3)}\iff \exists b(x)|f(x)=(x-3)b(x).$$
So, if $f$ exists then it must be 
$$(x-1)a(x)+1=(x-3)b(x).$$ We have for $x=1:$
$$1=-2b(1)\implies b(1)=-\dfrac 12,$$ which is not possible, since $b(x)$ is a polynomial with integer coefficientes and $1\in\mathbb{Z}$. (Note taht in a similar way we have $a(3)=-\frac12$.) Thus we conclude that there is no a polynomial with integer coefficients satisfying the given conditions. 
As it was said in comments your proof is essentially correct. If $f$ exists then it must be 
$$f(5)=4a(5)+1\equiv 1\bmod{4}$$ and $$f(5)=2b(5)\equiv 0\bmod{2}.$$ Since both congruences can't hold together you can conclude that the polynomial doesn't exist.
A: Your solution is short and correct. Indeed:
$$\begin{align}\begin{cases}f(x)\equiv 1\bmod{(x-1)}\\ f(x)\equiv 0\bmod{(x-3)}\end{cases} &\iff
\begin{cases}f(x)=(x-1)a(x)+1\\ f(x)=(x-3)b(x) \end{cases} \Rightarrow \\
\begin{cases}f(5)\equiv 1\bmod{4}\\f(5)\equiv 0\bmod{2}\end{cases} &\iff\begin{cases} f(5)=4a(5)+1\\ f(5)=2b(5) \end{cases}\\
0\equiv 1\bmod 2&\iff 0=2(2a(5)-b(5))+1 \Rightarrow \\
\emptyset &\iff \emptyset
\end{align}$$
