IMC 2012 · Day 1 · P5

First let us consider the case $a = 0$. The roots of $X^{2^{n+1}} + 1$ are exactly all primitive roots of unity of order $2^{n+2}$, namely $e^{2\pi i \frac{k}{2^{n+2}}}$ for odd $k = 1, 3, 5, \dots, 2^{n+2} - 1$. It is a cyclotomic polynomial, hence irreducible in $\mathbb{Q}[X]$.

Let now $a \ne 0$ and suppose that the polynomial in the question is reducible. Substituting $X = Y - \frac{a}{2}$ we get a polynomial $\left( Y - \frac{a}{2} \right)^{2^n} \left( Y + \frac{a}{2} \right)^{2^n} + 1 = \left( Y^2 - \frac{a^2}{4} \right)^{2^n} + 1$. It is again a cyclotomic polynomial in the variable $Z = Y^2 - \frac{a^2}{4}$, and therefore it is not divisible by any polynomial in $Y^2$ with rational coefficients. Let us write this polynomial as the product of irreducible monic polynomials in $Y$ with appropriate multiplicities, i.e. $$\left( Y^2 - \frac{a^2}{4} \right)^{2^n} + 1 = \prod_{i=1}^{r} f_i(Y)^{m_i} \quad f_i \text{ monic, irreducible, all different.}$$ Since the left-hand side is a polynomial in $Y^2$ we must have $\prod_i f_i(Y)^{m_i} = \prod_i f_i(-Y)^{m_i}$. By the above argument non of the $f_i$ is a polynomial in $Y^2$, i.e.\ $f_i(-Y) \ne f_i(Y)$. Therefore for every $i$ there is $i' \ne i$ such that $f_i(-Y) = \pm f_{i'}(Y)$. In particular $r$ is even and irreducible factors $f_i$ split into pairs. Let us renumber them so that $f_1, \dots, f_{\frac{r}{2}}$ belong to different pairs and we have $f_{i + \frac{r}{2}}(-Y) = \pm f_i(Y)$. Consider the polynomial $f(Y) = \prod_{i=1}^{r/2} f_i(Y)^{m_i}$. This polynomial is monic of degree $2^n$ and $\left( Y^2 - \frac{a^2}{4} \right)^{2^n} + 1 = f(Y) f(-Y)$. Let us write $f(Y) = Y^{2^n} + \dots + b$ where $b \in \mathbb{Q}$ is the constant term, i.e. $b = f(0)$.

Comparing constant terms we then get $\left( \frac{a}{2} \right)^{2^{n+1}} + 1 = b^2$. Denote $c = \left( \frac{a}{2} \right)^{2^{n-1}}$. This is a nonzero rational number and we have $c^4 + 1 = b^2$.

It remains to show that there are no rational solutions $c, b \in \mathbb{Q}$ to the equation $c^4 + 1 = b^2$ with $c \ne 0$ which will contradict our assumption that the polynomial under consideration is reducible. Suppose there is a solution. Without loss of generality we can assume that $c, b > 0$. Write $c = \frac{u}{v}$ with $u$ and $v$ coprime positive integers. Then $u^4 + v^4 = (b v^2)^2$. Let us denote $w = b v^2$, this must be a positive integer too since $u, v$ are positive integers. Let us show that the set $T = \{ (u, v, w) \in \mathbb{N}^3 \mid u^4 + v^4 = w^2 \text{ and } u, v, w \ge 1 \}$ is empty. Suppose the contrary and consider some triple $(u, v, w) \in T$ such that $w$ is minimal. Without loss of generality, we may assume that $u$ is odd. $(u^2, v^2, w)$ is a primitive Pythagorean triple and thus there exist relatively prime integers $d > e \ge 1$ such that $u^2 = d^2 - e^2$, $v^2 = 2de$ and $w = d^2 + e^2$. In particular, considering the equation $u^2 = d^2 - e^2$ in $\mathbb{Z}/4\mathbb{Z}$ proves that $d$ is odd and $e$ is even. Therefore, we can write $d = f^2$ and $e = 2g^2$. Moreover, since $u^2 + e^2 = d^2$, $(u, e, d)$ is also a primitive Pythagorean triple: there exist relatively prime integers $h > i \ge 1$ such that $u = h^2 - i^2$, $e = 2hi = 2g^2$ and $d = h^2 + i^2$. Once again, we can write $h = k^2$ and $i = l^2$, so that we obtain the relation $f^2 = d = h^2 + i^2 = k^4 + l^4$ and $(k, l, f) \in T$. Then, the inequality $w > d^2 = f^4 \ge f$ contradicts the minimality of $w$.

Remark 1. One can also use Galois theory arguments in order to solve this question. Let us denote the polynomial in the question by $P(X) = X^{2^n} (X + a)^{2^n} + 1$ and we will also need the cyclotomic polynomial $T(X) = X^{2^n} + 1$. As we already said, when $a = 0$ then $P(X)$ is itself cyclotomic and hence irreducible. Let now $a \ne 0$ and $x$ be any complex root of $P(x) = 0$. Then $\zeta = x(x + a)$ satisfies $T(\zeta) = 0$, hence it is a primitive root of unity of order $2^{n+1}$. The field $\mathbb{Q}[x]$ is then an extension of $\mathbb{Q}[\zeta]$. The latter field is cyclotomic and its degree over $\mathbb{Q}$ is $\dim_{\mathbb{Q}} \bigl( \mathbb{Q}[\zeta] \bigr) = 2^n$. Since the polynomial in the question has degree $2^{n+1}$ we see that it is reducible if and only if the above mentioned extension is trivial, i.e. $\mathbb{Q}[x] = \mathbb{Q}[\zeta]$. For the sake of contradiction we will now assume that this is indeed the case. Let $S(X)$ be the minimal polynomial of $x$ over $\mathbb{Q}$. The degree of $S$ is then $2^n$ and we can number its roots by odd numbers in the set $I = \{1, 3, \dots, 2^{n+1} - 1\}$ so that $S(X) = \prod_{k \in I} (X - x_k)$ and $x_k (x_k + a) = \zeta^k$ because Galois automorphisms of $\mathbb{Q}[\zeta]$ map $\zeta$ to $\zeta^k$, $k \in I$. Then one has $$S(X) S(-a - X) = \prod_{k \in I} (X - x_k)(-a - X - x_k) = (-1)^{|I|} \prod_{k \in I} \Bigl( X(X + a) - \zeta^k \Bigr) = T \bigl( X(X + a) \bigr) = P(X).$$ In particular $P\left( -\frac{a}{2} \right) = S\left( -\frac{a}{2} \right)^2$, i.e.\ $\left( \frac{a}{2} \right)^{2^{n+1}} + 1 = \left( \left( \frac{a}{2} \right)^{2^n} + 1 \right)^2$. Therefore the rational numbers $c = \left( \frac{a}{2} \right)^{2^{n-1}} \ne 0$ and $b = \left( \frac{a}{2} \right)^{2^n} + 1$ satisfy $c^4 + 1 = b^2$ which is a contradiction as it was shown in the first proof.

Remark 2. It is well-known that the Diophantine equation $x^4 + y^4 = z^2$ has only trivial solutions (i.e. with $x = 0$ or $y = 0$). This implies immediately that $c^4 + 1 = b^2$ has no rational solution with nonzero $c$.