PolynomialAndEquation IMO Difficult
1993


Problem - 2678
Show that $x^n + 5x^{n-1} + 3 = 0$ cannot be factorized into two non-constant polynomials with integer coefficients.

The notation $[x^t]P(x)$ denotes the coefficient of $x^t$ in polynomial $P(x)$. Suppose that $f(x)=g(x)h(x)$, where $g(x),h(x)$ are monic polynomials, $g(x),h(x)∈Z[x]$ and $deg(g) > 0$, $deg(h) > 0$, $deg(g) + deg(h)=n$. Let $g(x)=\displaystyle\sum_k\alpha_kx^k$, $h(x)=\displaystyle\sum_k\beta_kx^k$, where $\alpha_k,\beta_k$ are integers whenever $k \ge 0$. First, we have $\alpha_0\beta_0=3$, so $\mid\alpha_0\mid=3$ and $\mid\beta_0\mid=1$ or the other way around. WLOG, suppose that $\mid\alpha_0 =3$ and $\mid\beta_0\mid=1$. Let $m$ is the smallest positive integer such that $3\not\mid\alpha_m$ (such $m$ exists because $g(x)$ is monic). Now we have $[x^m]f(x)= \alpha_0\beta_m+\alpha_1\beta_{m−1}+\cdots+\alpha_m\beta_0\equiv \alpha_m \beta_0 \not\equiv \pmod{3}$. Therefore $m \ge n-1$, and $deg(g)\ge n−1$, thus $deg(h)\le 1$, hence $deg(h)=1$, and $h(x)=x+\beta_0$, here $\beta_0=\pm 1$, so $h(\pm 1)=0$, and $f(\pm 1)=0$. But it's obvious that $f(\pm 1)\ne 0$.

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