Saudi Arabia Mathematical Competitions

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

Observe that for $0\le k<n$, $$f(-2k)=(-1{)}^k(2k-1)(2k-3)\dots 1\cdot 3\dots (-2k+2n-1)$$ and $$\begin{array}{c}f(-2k-2)=f(-2(k+1))\\ =(-1{)}^{k+1}(2k+1)(2k-1)\dots 1\cdot 3\dots (-2k+2n-3).\end{array}$$ 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 $k=0,1,\dots ,n-1$. 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 $$a(a+2)(a+4)\dots (a+2n)+(a+1)(a+3)\dots (a+2n+1)=0,$$ 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.