XXIV OM - I - Problem 8

Find a polynomial with integer coefficients of the lowest possible degree, for which $ \sqrt{2} + \sqrt{3} $ is a root.
We will find a polynomial with integer coefficients, of which the root is the number $ a = \sqrt{2} + \sqrt{3} $. It is, of course, a root of the polynomial






This polynomial, however, does not have integer coefficients. Let's multiply the polynomial (1) by $ x - \sqrt{2} + \sqrt{3} $. We get $ (x - \sqrt{2})^2 - 3 $, that is,






Of course, the number $ \alpha $ is a root of the polynomial (2). However, not all of its coefficients are integers. Let's then multiply the polynomial (2) by $ x^2 + 2\sqrt{2}x - 1 $. We get,






The number $ \alpha $ is a root of the polynomial (3), which has integer coefficients.
We will prove that (3) is the polynomial of the lowest degree among such polynomials with integer coefficients, of which the root is the number $ \alpha $.
First, we will show that the polynomial (3) is not the product of two polynomials of positive degree with rational coefficients. If the polynomial (3) were divisible by a polynomial of the first degree with rational coefficients, then it would have a root in the set of rational numbers. Any such root is an integer that divides the constant term of the polynomial (3). It is easy to check that neither $ 1 $ nor $ - 1 $ is a root of this polynomial.
Next, we will prove that the polynomial (3) is not divisible by any quadratic polynomial with rational coefficients. If,






where $ a, b, c, d \in \mathbb{Q} $, then by comparing the coefficients of the same powers of $ x $ on both sides of equation (7), we would get the system of equations,






Solving for $ c $ from the first equation and for $ d $ from the last equation, and substituting into the remaining equations, we get,






Hence, $ a = 0 $, or $ b = \pm 1 $.
If $ a = 0 $, then $ b + \frac{1}{b} = - 10 $, that is, $ b^2 - 10 b + 1 = 0 $. This last equation has no roots in the set of rational numbers.
If $ b = 1 $, then $ a^2 = 12 $, and if $ b = - 1 $, then $ a^2 = 8 $. In neither of these cases is $ a $ a rational number.
If the number $ \sqrt{2} + \sqrt{3} $ were a root of some non-zero polynomial $ f $ of degree at most three with rational coefficients and this degree were minimal, then denoting by $ g $ and $ r $ the quotient and remainder of the division of the polynomial (3) by $ f $, we would get,






where $ \mathrm{deg} \, r < \mathrm{deg} \, f $. Here, $ r $ is not the zero polynomial, since, as we have shown, the polynomial (3) is not the product of two polynomials of positive degree with rational coefficients.
Substituting $ x = \sqrt{2} + \sqrt{3} $ in the last equation, we get $ r( \sqrt{2} + \sqrt{3}) = 0 $. However, the polynomial $ f $ had the smallest degree among non-zero polynomials with rational coefficients, of which the root is the number $ \sqrt{2} + \sqrt{3} $. Since $ \mathrm{deg} \, r < \mathrm{deg} \, f $, we have reached a contradiction.