Let $f(x) \in \mathbb Z[x]$ be an irreducible monic polynomial such that $|f(0)|$ is not a perfect square . Then is $f(x^2)$ also irreducible in $\mathbb Z[x]$ ?
( It is supposed to have an elementary solution , without using any field-extension etc. )
On
Let $\alpha$ be a root of $f$. Let $K = \mathbb{Q}(\alpha)$ and $L = \mathbb{Q}(\sqrt{\alpha})$. We have tower of extension $$L \supseteq K \supseteq \mathbb{Q}$$ By irreducibility of $f$, we know that $[K : \mathbb{Q}] = \deg f$ and obviously $[L : K] \leq 2$. If $[L : K] = 2$ then $[L : \mathbb{Q}] = 2 \deg f$ so that $f(x^2)$ must be irreducible for it must then be minimal polynomial for $\sqrt{\alpha}$. If $[L : K] = 1$ then obviously, $f(x^2)$ is reducible for it is divisible by minimal polynomial say $g(x) \in \mathbb{Q}[x]$ for $\sqrt{\alpha}$, which is of the same degree as $f$ because $\deg g = [L : \mathbb{Q}] = \deg f$. So whether $f(x^2)$ is irreducible or not is entirely decided by whether $L \not= K$ (equivalently, $\sqrt{\alpha} \in K$) or not.
Now if $L = K$ i.e. $f(x^2)$ is reducible then we further have factorization $f(x^2) = (-1)^{\deg g} g(x) g(-x)$ just by noting that if $\gamma$ is a root of $g$ then $-\gamma$ is also a root of $f(x^2)$ and hence, $|f(0)| = |g(0)|^2$ is a square. So if we assume $|f(0)|$ is not a square then $f(x^2)$ must be irreducible.
Sophisticated answer: Let us suppose $f(x)\in \mathbb{Z}[x]$ is monic and irreducible of degree $n$ but $g(x)=f(x^2)$ is reducible. Let $K\supset\mathbb{Q}$ be a splitting field of $g$ and let $a_1,\dots,a_n\in K$ be the roots of $f$, so $\pm\sqrt{a_1}\dots,\pm\sqrt{a_n}$ are the roots of $g$. Since $f$ is irreducible, $G=Gal(K/\mathbb{Q})$ acts transitively on the set $\{a_1,\dots,a_n\}$. It follows that the $G$-orbit of any root of $g$ must contain at least one of the square roots of $a_i$ for each $i$. Since $g$ is reducible, $G$ does not act transitively on the roots of $g$, so there must be exactly two orbits of roots of $g$, with each one containing exactly one of the square roots of each $a_i$. This means that $g$ has two irreducible factors $h_1$ and $h_2$ of degree $n$, and the roots of $h_1$ are the negatives of the roots of $h_2$. Thus the constant term of $h_1$ is $(-1)^n$ times the constant term of $h_2$ (if we take $h_1$ and $h_2$ to be monic). We conclude that the constant term of $g$ is $(-1)^n$ times the square of the constant term of $h_1$. It follows that $|f(0)|$ is a square.
(Strictly speaking, there is a special case when $a_i=0$ for some $i$, in which case $g$ has a double root and the argument above doesn't work as stated. But if that happens, then $|f(0)|=0$ is trivially a square. Also, the only $f$ for which this happens is $f(x)=x$)
Less sophisticated answer: Note that since $g(x)=g(-x)$, if $h(x)$ is an irreducible factor of $g(x)$, then so is $h(-x)$. So unless $h(x)=\pm h(-x)$ for some irreducible factor $h$ of $g$, the irreducible factors of $g$ come in pairs which have (up to sign) the same constant term, and so multiplying them all together we get that the constant term of $g$ is a square (up to sign).
So we just have to rule out $h(x)=\pm h(-x)$ for an irreducible monic factor $h$ of $g$. If $h(x)=-h(-x)$, then every term of $h$ has odd degree, and so $h$ is divisible by $x$. Thus $g(x)$ is divisible by $x$, and in particular its constant term is $0$, which is a square. If $h(x)=h(-x)$, then every term of $h$ has even degree, and we can write $h(x)=h'(x^2)$ for some $h'$. Writing $g(x)=h(x)k(x)$, we see that $k(x)$ also satisfies $k(x)=k(-x)$ (since $g$ and $h$ do), so $k(x)=k'(x^2)$ for some $k'$. But then $f(x)=h'(x)k'(x)$ is a factorization of $f$. This means that actually $k'(x)=1$, so $h'=f$ and $h=g$, so in this case $g$ is irreducible.