<s>[INST] Is $x^{2k+1} - 7x^2 + 1$ irreducible? 
Question. Is the polynomial $x^{2k+1} - 7x^2 + 1$ irreducible over $\mathbb{Q}$ for every positive integer $k$?

It is irreducible for all positive integers $k \leq 800$.
 [/INST] Here is a proof, based on a trick that can be used to prove that
$x^n + x + 1$ is irreducible when $n \not\equiv 2 \bmod 3$.
We work with Laurent polynomials in $R = \mathbb Z[x,x^{-1}]$; note that
$R$ has unit group $R^\times = \pm x^{\mathbb Z}$.
We observe that for $f \in R$, the sum of the squares of the coefficients
is given by
$$\|f\| = \int_0^1 |f(e^{2 \pi i t})|^2\,dt
        = \int_0^1 f(e^{2 \pi i t}) f(e^{-2 \pi i t})\,dt
        = \int_0^1 f(x) f(x^{-1})\big|_{x = e^{2 \pi i t}}\,dt .$$
Now assume that $f(x) = g(x) h(x)$. Then, since
$f(x) f(x^{-1}) = \bigl(g(x)h(x^{-1})\bigr)\bigl(g(x^{-1})h(x)\bigr)$,
$G(x) := g(x) h(x^{-1})$ satisfies $\|G\| = \|f\|$
and $G(x) G(x^{-1}) = f(x) f(x^{-1})$.
Now we consider $f(x) = x^n - 7 x^2 + 1$; then $\|f\| = 51$.
If $f = g h$ as above, then write
$G(x) = \pm x^m(1 + a_1 x + a_2 x^2 + \ldots)$
and $G(x^{-1}) = \pm x^l(1 + b_1 x + b_2 x^2 + \ldots)$.
The relation $G(x) G(x^{-1}) = f(x) f(x^{-1})$ translates into
(equality of signs and)
$$(1 + a_1 x + \ldots)(1 + b_1 x + \ldots) = 1 - 7 x^2 + O(x^{n-2}).$$
Assuming that $n > 40$ and considering terms up to $x^{20}$, one can
check (see below) that the only solution such that
$a_1^2 + a_2^2 + \ldots + a_{20}^2 + b_1^2 + b_2^2 + \ldots + b_{20}^2\le 49$
is, up to the substitution $x \leftarrow x^{-1}$, given by
$1 + a_1 x + \ldots = 1 - 7x^2 + O(x^{21})$,
$1 + b_1 x + \ldots = 1 + O(x^{21})$. Since the $-7$ (together with the
leading and trailing 1) exhausts our allowance
for the sum of squares of the coefficients, all other coefficients
must be zero, and we obtain that $G(x) = \pm x^a f(x)$
or $G(x) = \pm x^a f(x^{-1})$. Modulo interchanging $g$ and $h$,
we can assume that $g(x) h(x^{-1}) = \pm x^a f(x)$,
so $h(x^{-1}) = \pm x^a f(x)/g(x) = \pm x^a h(x)$,
and $x^{\deg h} h(x^{-1})$ divides $f(x)$.
This implies that $h(x)$ divides $x^n f(x^{-1})$, so $h(x)$ must divide
$$f(x) - x^n f(x^{-1}) = 7 x^2 (x^{n-4} - 1).$$
So $h$ also divides
$$f(x) - x^4 (x^{n-4} - 1) = x^4 - 7 x^2 + 1 = (x^2-3x+1)(x^2+3x+1).$$
Since $h$ also divides $f$, it must divide the difference $x^n - x^4$,
which for $n \neq 4$ it clearly doesn't, since the quartic has no
roots of absolute value 0 or 1; contradiction.
The argument shows that $x^n - 7 x^2 + 1$ is irreducible for $n > 40$;
for smaller $n$, we can ask the Computer Algebra System we trust.
This gives:
Theorem. $x^n - 7 x^2 + 1$ is irreducible over $\mathbb Q$
for all positive integers $n$ except $n=4$.
ADDED 2017-01-08:
After re-checking the computations, I realized that there
was a small mistake that ruled out some partial solutions prematurely.
It looks like one needs to consider terms up to $x^{20}$.
Here is a file with MAGMA code that verifies that
there are no other solutions.
</s>