http://www.kalva.demon.co.uk/usa/usoln/usol893.html Alternate solution : Let $r_1, r_2, \ldots, r_n$ be the real roots of the polynom $P$ in $\mathbb{C}$. Thus, we may write : \[ P(i) = (i - r_1)(i - r_2) \ldots (i - r_n), \] hence, \[ |i - r_1| \cdot |i - r_2| \cdot \ldots \cdot |i - r_n| < 1. \] Since $P$ has real coefficients, its non-real roots can be pairwise gathered, each pair being constituted of two conjugate numbers. But, if $r \in \mathbb{R}$, $|i - r| = \sqrt{1 + r^2} \ge 1$. So, there exist two reals $a$ and $b$ such that $P(a + ib) = 0$ and \[ 1 > |i - (a + ib)| \cdot |i - (a - ib)| = \sqrt{(a^2 + b^2 + 1)^2 - 4b^2}, \] hence : \[ 4b^2 + 1 > (a^2 + b^2 + 1)^2. \]