XVII OM - I - Problem 1

Present the polynomial $ x^5 + x + 1 $ as a product of two polynomials of lower degree with integer coefficients.
Suppose the desired factorization exists. None of the factors of this factorization can be of the first degree. For if for every $x$ the equality held




where the numbers $m, n, a, b, \ldots$ are integers, then it would follow that $ma = 1$, so $m = a = \pm 1$, hence the polynomial $x^5 + x + 1$ would have an integer root $-\frac{n}{m}$. This is impossible, however, because for every integer $x$ the number $x^5 + x + 1$ is odd.
Hence, the desired factorization must have the form




where the numbers $m, n, p, a, b, c, d$ are integers.
In this case, $ma = 1$, $pd = 1$, so $m = a = \pm 1$, $p = d = \pm 1$.
Thus, there are four possible cases, but it suffices to consider two of them; the remaining cases reduce to these by changing the sign of both factors in the factorization.
a) When $m = a = 1$, $p = d = 1$, then by comparing the coefficients of like powers of $x$ on both sides of the equality




we obtain




The system of equations (1) - (4) has one solution $n = 1$, $b = -1$, $c = 0$, which gives the factorization




b) When $m = a = 1$, $p = d = -1$, we similarly obtain the system of equations




It is easy to check that this system has no solutions. The solution to the problem is therefore the formula




where $\varepsilon = \pm 1$.
Note. The factorization sought in the problem can be found quickly in the following way: 




Such a solution to the problem still requires a proof that there is no other factorization besides the one obtained by changing the signs of both found factors. The easiest way to show this is by using complex numbers. Let $x_1$, $x_2$ denote the roots of the polynomial $x^2 + x + 1$, and $x_3$, $x_4$, $x_5$ the roots of the polynomial $x^3 - x^2 + 1$.
Then




Any quadratic divisor of the polynomial $x^5 + x + 1$ with integer coefficients is of the form $m(x - x_i)(x - x_k)$, where $m$ is an integer and $x_i$, $x_k$ are two roots of the polynomial. It is easy to see that such a divisor can only be $m(x - x_1)(x - x_2)$. The detailed proof of this is left as an exercise.
Hint. One of the numbers $x_3$, $x_4$, $x_5$ is an irrational real number, and the other two are complex conjugates, whose sum and product are irrational.