Let $f(x)=(x-a_1^2)(x-a_2^2)\ldots(x-a_n^2)$, where $a_i$ are complex numbers ($a_i^2$ are all distinct, since they are the roots of an irreducible polynomial). Now assume that $f(x^2)=g_1(x)\cdot\ldots\cdot g_k(x),\ k\ge 2$ is a decomposition of $f(x^2)$ into $>1$ irreducible monic integral polynomials. If for some $i$ there is a $j$ s.t. both $a_j,-a_j$ are roots of $g_i$, then for any $p$ s.t. $a_p$ is a root of $g_i,\ -a_p$ will also be a root of $g_i$. This is because $a_p$ has both $a_j,-a_j$ as conjugates, which means that $-a_p$ also has them as conjugates, so $-a_p$ can't appear as a root in any $g$ except for $g_i$. This means that $g$ is a product of expressions of the form $x^2-a_i^2$, which makes it a factor of $f$, so we get a contradiction. The above implies that if $a_j$ is a root of $g_i$, then $-a_j$ is not. Now, if $g_i$ has the distinct roots $a_{i_1},\ldots,a_{i_t}$, then there is another $g_{i'}$ which has roots $-a_{i_1},\ldots,-a_{i_t}$. The $g$'s thus come in pairs, each two members of the same pair having the same absolute value of the free term. When we multiply all these free terms together, we find that the absolute value of the free term of $f$ is a square, thus getting another contradiction.

Very nice result Here are two different proofs,the second is the official solution.

hxy wrote: Very nice result Here are two different proofs,the second is the official solution. Can you repost your two solutions? I can't read it @grobber: I don't understand your solution, may be it false! Please explain more, sorry if I false!

Something to note: the leading coefficient of $1$ part is totally unnecessary (which the three posted solutions don't even use actually!). Here is a proof which is essentially rigorozing grobber's arguments. IMO it is more intuitive than hxy's solution because the concept of conjugation via automorphisms is very core to irreducible polynomial theory. Let $f(x) = c(x-a_1^2) \cdot ... \cdot (x-a_n^2)$ for distinct complex numbers $a_1,...,a_n$ and $c$ an integer. Then suppose $f(x^2) = g_1(x) \cdot ... \cdot g_m(x)$ for irreducible integer polynomials $g_i(x)$. Note that $\pm a_i$ for $1 \le i \le i$ are the roots of $f$. Suppose $a_i, -a_i$ are both roots of $g_j(x)$ for some $j$. Take the splitting field of $f(x^2)$, take an automorphism $A$ which maps $a_i^2$ to $a_k^2$ (which exists since they are the roots of the same minimum polynomial). But then $A(a_i) = \pm a_k$. But then $\{A(a_i), A(-a_i)\} = \{a_k, -a_k\}$. It follows that $g_j$'s roots are $\pm a_1, \pm a_2, ..., \pm a_n$ because automorphisms of a number map to another root of the minimum polynomial. But then $f(x^2)$ is irreducible, so we'd be done. Therefore assume $a_i, -a_i$ are never the roots of the same minimum polynomial. Now plug $-x$ into $f(x^2) = g_1(x) \cdot ... \cdot g_m(x)$. We easily arrive at that for each $i$, there exists some $j \neq i$ such that $g_i(x) = \pm g_j(-x)$ (remark that the $g_i$ are not necessarily distinct so $j$ is not unique, however we don't care about this). But then by pairing the product $g_1(x) \cdot ... \cdot g_m(x)$ into products $\pm g_i(x)g_j(-x)$ we get the constant term of $f(x^2)$ is equal in magnitude to a product of perfect squares, implying $|f(0)|$ is a perfect square which is absurd so we are done. Remark: By considering $x^{\deg f}f(1/x)$, we can strengthen this to either $|f(0)|$ is not a perfect square or the absolute value of $f$'s leading coefficient is not a perfect square.

dinoboy wrote: Something to note: the leading coefficient of $1$ part is totally unnecessary. Wrong. Take $f(x) = 3x^2+12$; then $f(x)$ is irreducible in $\mathbb{Z}[x]$ and $|f(0)| = 12$ is not a perfect square. However, $f(x^2) = 3x^4+12 = 3(x^2-2x+2)(x^2+2x+2)$, so $f(x^2)$ is not irreducible. And yes, Lumpa, your proof is correct (albeit a tad unkempt).

A more straightforward solution: Assume $f(x^2)=g(x)h(x)$ for some nonconstant $g$ and $h$ in $Z[x]$. Then $f(x^2)=g(-x)h(-x)$. Multiplying these two we have: $f(x^2)^2 = g(x)g(-x)h(x)h(-x)$ However $g(x)$$g(-x)$ and $h(x)$$h(-x)$ are even polynomials so for some $r(x)$ and $s(x)$ we have $g(x)g(-x) = r(x^2)$ and $h(x)h(-x) = s(x^2)$ So $f(x^2)^2=r(x^2)s(x^2)$ so $f(x)^2=r(x)s(x)$ However $f$ is irreducible and monic so either $r(x) = s(x) = f(x)$ or $r(x) = s(x) = -f(x)$ In the first case we have $f(x^2)=g(x)h(x)=r(x^2)=g(x)g(-x)$ so $h(x)=g(-x)$ so $h(0)g(0)=h(0)^2=f(0)$, contradiction. In second case $f(x^2)=g(x)h(x)=-r(x^2)=-g(x)g(-x)$ so $h(x)=-g(x)$ so $h(0)g(0)=-h(0)^2=f(0)$, contradiction. Q.E.D

erfankhayyam wrote: So $f(x^2)^2=r(x^2)s(x^2)$ so $f(x)^2=r(x)s(x)$ However $f$ is irreducible and monic so either $r(x) = s(x) = f(x)$ or $r(x) = s(x) = -f(x)$ Sorry, but why is this true? Thanks!

@MathPanda1: If $f(x^2)^2=r(x^2)s(x^2)$ for all real $x$, $f(t)^2=r(t)s(t)$ for all positive $t$, using the obvious substitution. Equality of the functions at infinitely many points implies equality of the polynomials, and $f(x)^2=r(x)s(x)$. Then, from the irreducibility of $f$, $f^2(x)$ can only factor as $(af^2(x))\cdot b$, $(af(x))\cdot (bf(x))$, or $a\cdot (bf^2(x))$ with $a$ and $b$ constants. The first and third possibilities are excluded by prior hypothesis, leaving the second - and then $a$ and $b$ are restricted by the fact that $f$ is monic.

Suppose $f(x^2)$ is not irreducible. Then there exists irreducible $g(x)$ with $0<\deg g < 2\deg f$ with $g(x)\mid f(x^2)$, which implies $g(-x)\mid f(x^2)$ as well. If $g(x)=g(-x)$, then $g(x)=g_0(x^2)$ for some $g_0$, implying $g_0(x)\mid f(x)$, a contradiction. Thus $g(x)\neq g(-x)$, and as both are irreducible, they are relatively prime. So \[ g(x)g(-x)\mid f(x^2). \]Let $g(x)g(-x)=g_1(x^2)$, with \[ \deg f(x^2)\ge\deg g_1(x^2)=2\deg g(x). \]Now, if $\deg g< \deg f$, then $g_1(x)\mid f(x)$, again a contradiction. So $\deg g =\deg f$, and as $f$ is monic, \[ g(x)g(-x)=\pm f(x^2) \]But now plugging in $0$ yields \[ |f(0)|=g(0)^2 \]a contradiction.

jmerry wrote: @MathPanda1: If $f(x^2)^2=r(x^2)s(x^2)$ for all real $x$, $f(t)^2=r(t)s(t)$ for all positive $t$, using the obvious substitution. Equality of the functions at infinitely many points implies equality of the polynomials, and $f(x)^2=r(x)s(x)$. Then, from the irreducibility of $f$, $f^2(x)$ can only factor as $(af^2(x))\cdot b$, $(af(x))\cdot (bf(x))$, or $a\cdot (bf^2(x))$ with $a$ and $b$ constants. The first and third possibilities are excluded by prior hypothesis, leaving the second - and then $a$ and $b$ are restricted by the fact that $f$ is monic. But how do know if $f(x)=(x-r_1)(x-r_2)...(x-r_n)$, then $f(x)^2=r(x)s(x)$, for example, where $r(x)=(x-r_1)^2...(x-r_i)^2(x-r_{i+1})...(x-r_j)$ and $s(x)=(x-r_{j+1})^2...(x-r_k)^2(x-r_{k+1})...(x-r_n)$, where $r(x)$ and $s(x)$ have integer coefficients. How do you prove that this cannot be the case? Thanks!

@MathPanda1: $f$ is irreducible. That's why you can't do that. ... OK, I need to quote the unique factorization property for polynomials. We're trying to factor the square of a prime here; there just isn't much choice.

Am I missing something, or does this work? Quote: We will show the contrapositive; that if $P$ is a monic irreducible in $\mathbb Q[x]$ and $P(x^2)$ factors into monic irreducibles $\prod_{i\in I} P_i$, then $|P(0)|$ is a rational square. Choose $P_i$ of minimal degree, and take $\omega$ to be one of its roots. Observe that $\omega^2$ is then a root of $P$. Clearly $\omega^2\in\mathbb Q(\omega)$, and thus $$2\cdot\text{deg}(P)>\text{deg}(P_i)=[\mathbb Q(\omega):\mathbb Q]\geq [\mathbb Q(\omega^2):\mathbb Q]=\text{deg}(P)$$It follows that $[\mathbb Q(\omega):\mathbb Q]=[\mathbb Q(\omega^2):\mathbb Q]$ and thus that $\mathbb Q(\omega)\simeq \mathbb Q(\omega^2)$. There is then equality mediated through norms $$|P(0)|=|N_{\mathbb Q(\omega)/\mathbb Q}(\omega^2)|=|N_{\mathbb Q(\omega)/\mathbb Q}(\omega)|^2=|P_i(0)|^2$$Validating our earlier assertion $\blacksquare$

Let $f(x^2) = p(x) q(x)$, for some $p, q \in \mathbb{Z}[x].$ Assume for contradiction that $p, q$ are nonconstant. Then, we can decompose them as $p(x) = e_1(x) + o_1(x), q(x) = e_2(x) + o_2(x)$, where $e_1, e_2$ are both even polynomials and $o_1, o_2$ are both odd polynomials. Consider the decomposition where $p$ is of minimal degree. Since $e_1e_2, o_1o_2$ are both even, we must have $e_1 o_2 + e_2 o_1 = 0.$ Observe that $\gcd(e_1, o_1) = 1$, as otherwise there clearly exists a nonconstant factor of $p$ which has lower degree. Hence, there must exist a polynomial $f \in \mathbb{Q}[x]$ with $e_2 = f e_1, o_2 = -f o_1.$ We hence have that $f(x^2) = p(x) q(x) = f (e_1^2 - o_1^2).$ Since $e_1^2 - o_1^2$ has all exponents even, it can be written as $k(x^2)$ for some $k \in \mathbb{Z}.$ Furthermore, $f$ must also be expressible as $\ell(x^2)$ for some $\ell \in \mathbb{Q}[x].$ This implies that $f(x) = \ell k$, for $\ell, k \in \mathbb{Q}[x].$ Hence, because $f$ is irreducible, we know that $\ell$ is a constant polynomial. Because the leading coefficient and constant term of $e_1^2-o_1^2$ are both squares or negative squares, while the leading coefficient of $f(x)$ is one, $f(x) = \ell k$ clearly implies that the constant term of $f(x)$ is a square. This is absurd. $\square$

Suppose otherwise that $f(x^2) = g(x)\cdot h(x)$ is reducible in $\mathbb{Z}[x]$. Then we have \[ g(x)h(x) = f(x^2) =g(-x)h(-x) \]$\textbf{Claim 01.}$ $\gcd(h(x),h(-x)) \not= 1$. $\textit{Proof.}$ Suppose otherwise, that $\gcd(h(x),h(-x)) = 1$. Then we would have $h(x) | g(-x)$, which implies $g(-x) = h(x)c(x)$ where $c \in \mathbb{Z}[x]$. Keep in mind that $f$ is monic. Now, we have $c(x) = c(-x)$, which implies there exists a polynomial $d \in \mathbb{Z}[x]$ such that \[ c(x) = d(x^2) \]Therefore, $d(x^2) | f(x^2)$. If $d$ is the constant polynomial, this forces $d \equiv 1$. But $|f(0)| = |g(0)h(0)| = |h(0)|^2$, a contradiction. Therefore, $d(x^2) | f(x^2)$ for all $x \in \mathbb{Z}$. Since there are infinitely many such $x$ satisfying such property. This forces $d(x) | f(x)$. But we know that $f$ is irreducible, forcing $d \equiv \pm f$, a contradiction. Therefore, we have $\gcd(h(x), h(-x)) \not= 1$ for all such polynomials $h$. The trick now is to consider such minimal factor of $f(x^2)$ since we assume that $f(x^2)$ is reducible. This factor must have degree greater than $1$. Since $\gcd(h(x), h(-x)) \not= 1$. This must be a polynomial of degree greater than $1$. But if this isn't $h(x)$, this contradicts the assumption of minimality of $h(x)$. This forces $\gcd(h(x), h(-x)) = h(x)$. Hence $h(-x) | h(x)$. Both $h(-x)$ and $h(x)$ are polynomials with the same degree and coefficients being $1$ or $-1$ since $f$ is monic. This gives us 2 choices : If $h(x) = -h(-x)$. Plug $x = 0$, and we get $h(0) = 0$. But, since $h(x) | f(x^2)$. Then $|f(0)| = 0$, which is a perfect square, contradiction. If $h(x) = h(-x)$. Then there must exists a polynomial $j$ such that $h(x) = j(x^2)$. Therefore, $j(x^2) | f(x^2)$ for infinitely many $x$, which forces $j(x) | f(x)$. Since $f$ is irreducible and $j$ is non constant, this gives us $j \equiv \pm f$, a contradiction as we must have $\text{deg} \ j < \text{deg} \ f$

We will prove the contrapositive: if $f(x^2)$ is reducible, then $|f(0)|$ is a perfect square. This solution hinges on the following claim. Claim. If $f(x^2)$ is divisible by an even polynomial $p(x^2)$, then either $p$ is constant or $f \equiv \pm p$. Proof. Suppose $f(x^2) = p(x^2)q(x)$ for some polynomial $q(x)\in \mathbb Z[x]$. Plugging in $x \mapsto -x$ and simplifying gives $q(x) = q(-x)$; this means $q(x) = q_0(x^2)$ for some polynomial $q_0(x)\in\mathbb Z[x]$. But then $f(x) = p(x)q_0(x)$ is a factorization of $f$, so either $p$ or $q_0$ is constant. This proves the claim. $\square$ Now take a factor $h(x)$ of $f(x^2)$. Then $h(-x)$ is also a factor of $f(x^2)$ by similar logic as above. Let $r(x) := \gcd(h(x),h(-x))$; then \[ h(x) = r(x)s(x)\quad\text{and}\quad h(-x) = r(x)s(-x) \]for some polynomial $s(x)$ with $\gcd(r(x),s(x)) = 1$. It follows that $r(x)\cdot [s(x)s(-x)]$ divides $f(x^2)$. But $s(x)s(-x)$ is an even factor of $f(x^2)$ with positive degree, so we actually obtain the factorization $f(x^2) = \pm s(x)s(-x)$. Now $|f(0)| = |s(0)|^2$ is a perfect square, which is what we were after.

If not, let $f(x^2)=g(x)h(x),$ where $g(x),h(x)\in\mathbb Z[x],\deg g,\deg h\geq 1.$ Define $g(\sqrt x)=g_1(x)+\sqrt x\cdot g_2(x),h(\sqrt x)=h_1(x)+\sqrt x\cdot h_2(x)$ where $g_1(x),g_2(x),h_1(x),h_2(x)\in\mathbb Z[x].$ Then $$\begin{aligned}f(x)&=g(\sqrt x)h(\sqrt x)=\left(g_1(x)+\sqrt x\cdot g_2(x)\right)\cdot\left(h_1(x)+\sqrt x\cdot h_2(x)\right)\\ &=g_1(x)h_1(x)+xg_2(x)h_2(x)+\sqrt x\left(g_2(x)h_1(x)+g_1(x)h_2(x)\right).\end{aligned}$$Then $g_2(x)h_1(x)+g_1(x)h_2(x)=0\Rightarrow g_2(x)h_1(x)=-g_1(x)h_2(x).$ It is well known that $\exists p(x),q(x),r(x),s(x)\in\mathbb Z[x],$ such that $$g_2(x)=p(x)q(x),h_1(x)=r(x)s(x),g_1(x)=p(x)r(x),h_1(x)=-g_1(x)h_2(x).$$Therefore $$\begin{aligned}f(x)&=g_1(x)h_1(x)+xg_2(x)h_2(x)\\ &=r^2(x)s(x)p(x)-xq^2(x)s(x)p(x)\\ &=s(x)p(x)\left(r^2(x)-x^2q^2(x)\right).\end{aligned}$$As the leading coefficient of $f{}$ is ${1}{},s(x),p(x)\equiv\pm 1.$ Then $$|f(x)|=\left|r^2(x)-x^2q^2(x)\right|.$$Therefore $|f(0)|=|r^2(x)|=|r(x)|^2,$ contradiction$!\blacksquare$

FTSOC suppose that $f(x^2) = g(x) \cdot h(x)$ for nontrivial monic $g$ and $h$. Note that $g$ and $h$ must have some odd power. Claim: $f(x^2) = m(x)m(-x)$ holds for no polynomial $m$. Proof. Plug in $x = 0$. $\blacksquare$ Claim: $f$ has no duplicate roots. Proof. Consider the derivative of $f$, that would give a smaller minimal polynomial. $\blacksquare$ Claim: For any root $z$, $g(\sqrt{z}) = 0 \iff g(-\sqrt{z}) = 0$. Proof. Suppose that FTSOC $g(\sqrt{z}) = h(-\sqrt{z}) = 0$. We can factor out the minimal polynomial $m$ of $\sqrt{z}$ to get \[ g(x)h(x) = g_1(x)m(x) \cdot h_1(x)m(-x). \]which implies that $m(x)m(-x)$ divides $f(x^2)$ nontrivially, which contradicts the irreducible of $f$. $\blacksquare$ However, then by pairing together roots $\sqrt{z}$ and $-\sqrt{z}$ we get \[ g(x) = \prod_{z \in S} (x^2 - z) \]for some set $S$. This contradicts the minimality of $f$.

Suppose that there exists a polynomial $f$ that contradicts the result. Then $f(x^2) = g(x)h(x)$ for some $g, h$. Suppose $\deg g < \deg f < \deg h$. Now consider the polynomial $g(x)g(-x) = p(x^2)$ for some polynomial $p$. For any $\beta^2 = \alpha$ such that $g(\beta) = 0$, it follows that $p(\alpha) = 0$. Furthermore, $\alpha$ is a root of $f$, but this contradicts minimality of $f$. Hence $\deg g = \deg f = \deg h$, and furthermore $g(x)g(-x) = f(x^2)$ exactly. It follows that $h(x) = g(-x)$, and $f(0) = g(0)^2$ is a perfect square.