Suppose $2k + 1 \equiv 3 \mod 4$ in $\Bbb{Z}_{\geq 1}$.
Is the polynomial: $p_k(x) = x^{2k + 1} + x^{2k - 1} \dots + x + 1$ irreducible in $\Bbb{Z}_2[x]$?
I do not know whether it is true or not...
(Side note: I've had a question in a test that is solved immediately if the statement above turns out to be true (a two parted question that would be solved by proving a single statement). The question was much simpler then trying to prove/disprove the above statement but I missed the easy way (very unfortunately and quite disappointingly after so much preparation and work). I decided to go about it my own way, brute forcing instead of trying to think outside the box.)
The idea I had in mind - the statement is true:
Suppose for the sake of contradiction that $p_k$ is reducible and suppose $g, f \in \Bbb{Z}_2[x]$ are such that $gf = p_k$ and $\deg(f), \deg(g) \geq 1$. Also w.l.o.g $\deg f > \deg g$.
Denote $$ f = x^s + b_{s - 1}x^{s - 1} + \dots + b_1 x + 1 \\ g = x^r + a_{r - 1}x^{r - 1} + \dots + a_1 x + 1 $$
Note then, that $r + s = 2k + 1$ so either $r$ is even and $s$ is odd or vice versa. Suppose $r$ is even and $s$ is odd.
We now proceed to equate the coefficients between $fg$ and $p_k$.
Also, $a_1 + b_1 \equiv 1 \mod 2$ because the coefficient of $x$ in $p_k$. Suppose $b_1 \equiv 0 \mod 2 \wedge a_1 \equiv 1 \mod 2$.
Now for the coefficient of $x^2$ we have $a_1b_1 + a_2 + b_2 \equiv a_2 + b_2 \mod 2 \equiv 0 \mod 2$. Yielding yet again two possibilities. Suppose $a_2 \equiv b_2 \mod 2 \equiv 1 \mod 2$.
And in general we can say: $$ a_l + b_l + \sum_{i = 1}^{l - 1}a_i b_{l - i} \equiv \cases{1 \mod 2 & l is odd \\ 0 \mod 2 & l is even} $$ Where $l \in \{1, 2, 3, \dots, r + s - 1\}$. I think maybe I need to prove that all the even powers must reside in $f$ where all the odd powers must reside in $g$ and get a contradiction in some way. But I really don't know what I should do. Maybe induction? But what is the hypothesis?
Thank you for your trouble! And I am sorry if what I've tried above is confusing and unclear. I had no time to think when I tried to solve it and I am kind of stuck now.