Saudi Arabian Mathematical Competitions

Consider the polynomial function
$$f(x) = x(x+2)(x+4) \ldots(x+2n) + (x+1)(x+3) \ldots(x+2n-1)$$
We have $\deg f = n+1$ and the leading coefficient is $1$.

Observe that for $0 \leq k < n$,
$$f(-2k) = (-1)^k (2k-1)(2k-3) \ldots 1 \cdot 3 \ldots (-2k+2n-1)$$
and
$$\begin{gathered} f(-2k-2) = f(-2(k+1)) \\ = (-1)^{k+1} (2k+1)(2k-1) \ldots 1 \cdot 3 \ldots (-2k+2n-3) . \end{gathered}$$
It is clear that $f(-2k-2) f(-2k) < 0$, that is, the interval $I_k = (-2k-2, -2k)$ contains a root of $f$ for any $k = 0, 1, \ldots, n-1$. Because $f$ is of degree $n+1$ and it has $n$ real roots, it follows that $f$ has $n+1$ real roots.

It is clear that $f$ has integer coefficients. If $f$ has a rational root $a$, then $a$ must be integer since the leading coefficient is $1$.

If $a$ is even, then from $f(a) = 0$ we get
$$a(a+2)(a+4) \ldots(a+2n) + (a+1)(a+3) \ldots(a+2n+1) = 0,$$
that is, an odd number is equal to $0$, not possible.

If $a$ is odd, then from $f(a) = 0$, it follows again that an odd number is equal to $0$, not possible.