, 2014

Unless otherwise stated, we work in $\mathbb{Q}[X]$. Since the case $\deg f = 1$ is easily dealt with, let $\deg f \ge 2$ and write $f(X^2) = gh$, where $g$ and $h$ both have a positive degree, and $g$ is irreducible. Next, write $g = a(X^2) + Xb(X^2)$ and $h = c(X^2) + Xd(X^2)$ to infer (from $f(X^2) = gh$ by an obvious argument on the parity of degrees) that
$$ad + bc = 0, \tag{1}$$
so $f = ac + Xbd$, whence
$$af = (a^2 - Xb^2)c. \tag{2}$$
We now show that $a$ and $b$ are coprime. Alternatively, but equivalently, $\delta = \gcd(a, b)$ is a constant. To this end, write $a = a_1\delta$ and $b = b_1\delta$, and refer to the irreducibility of $g$ to deduce that $\delta(X^2)$ is either associated with $g$, a case to be ruled out in the sequel, or a constant, in which case we are through.
In the former case, $a_1(X^2) + Xb_1(X^2)$ is a constant, so $b_1 = 0$, whence $b = 0$ and $g = a(X^2)$, and (1) forces one of $a$ and $d$ to be $0$. The fact that $g$ is not constant rules out the case $a = 0$, so $d = 0$, $f = ac$ and $h = c(X^2)$. Since $f$ is irreducible,

one of $a$ and $c$ must be a constant, hence so must be one of $g$ and $h$ — a contradiction, since both have a positive degree. Incidentally, notice that we have just proved that $b \neq 0$.

Notice further that $a$ and $X$ are also coprime: otherwise, $a(0) = 0$, so $g(0) = 0$, hence $f(0) = 0$, contradicting the fact that $f$ is irreducible and $\deg f \geq 2$.

Consequently, $a$ and $a^2 - Xb^2$ are coprime, so $a$ divides $c$ by (2), and $f = (a^2 - Xb^2)c_1$ for some $c_1$ in $\mathbb{Q}[X]$. Since $f$ is irreducible, one of $a^2 - Xb^2$ and $c_1$ must be a constant. Since $b \neq 0$ by the remark at the end of the last but one paragraph, $a^2 - Xb^2$ cannot be constant, so $c_1$ is a constant.

From now on we work in $\mathbb{Z}[X]$. By the preceding, $nf = m(u^2 - Xv^2)$ for some integers $m$ and $n$, and some $u$ and $v$ in $\mathbb{Z}[X]$. Fix a prime integer $p$ and write $m = p^\mu m_1$, $n = p^\nu n_1$, $u = p^\alpha u_1$, $v = p^\beta v_1$, where $\alpha, \beta, \mu, \nu$ are non-negative integers, and none of $m_1, n_1, u_1, v_1$ is divisible by $p$. To make a choice, let $\alpha \leq \beta$; the case $\alpha > \beta$ is dealt with similarly. Since $f$ is primitive, the relation
$$p^\nu n_1 f = p^{\mu+2\alpha} m_1 \left( u_1^2 - X p^{2(\beta-\alpha)} v_1^2 \right)$$
implies that $\nu \geq \mu + 2\alpha$. If we show that $\nu = \mu + 2\alpha$, we are through.

Suppose, if possible, that $\nu > \mu + 2\alpha$, to deduce that $p$ divides $u_1^2 - Xp^{2(\beta-\alpha)}v_1^2$, so it also divides $u_1^2(X^2) - X^2p^{2(\beta-\alpha)}v_1^2(X^2) = (u_1(X^2) - Xp^{\beta-\alpha}v_1(X^2))(u_1(X^2) + Xp^{\beta-\alpha}v_1(X^2))$. Since $p$ is prime, it must divide one of $u_1(X^2) \pm Xp^{\beta-\alpha}v_1(X^2)$, and an obvious argument on the parity of degrees shows that $u_1(X^2)$ must be divisible by $p$, and hence so must be $u_1$ — a contradiction which concludes the proof.