China Mathematical Competition (Complementary Test)

Let $$f(x)=(x+1)(x+2)\cdots (x+n)+2.\stackrel{◯}{1}$$ Obviously, $f(x)$ is a monic polynomial of degree $n$ with positive integer coefficients. We are going to prove that $f(x)$ has property (2). For any integer $t$, since $n\ge 4$, we know that there exists definitely a multiple of 4 in any $n$ consecutive numbers $t+1,t+2,\dots ,t+n$. Then from $\stackrel{◯}{1}$, we have $f(t)\equiv 2(\mathrm{mod}4)$. Then for any $k$ ($k\ge 2$) positive integers $r_1,r_2,\dots ,r_k$, we have $$f(r_1)f(r_2)\cdots f(r_k)\equiv 2^k\equiv 0(\mathrm{mod}4).$$ On the other hand, for any positive integer $m$, we have $f(m)\equiv 2(\mathrm{mod}4)$. Therefore, $$f(m)\not\equiv f(r_1)f(r_2)\cdots f(r_k)(\mathrm{mod}4),$$ which implies that $f(m)\ne f(r_1)f(r_2)\cdots f(r_k)$. We then find the required $f(x)$ and complete the proof. $\square$