Let $F_q$ be the finite field of $q$ elements. I am looking for the existence of an irreducible trinomial of the following form: $$x^n-x^m-1$$ over $F_q$ for some $n,m.$ I think it should be true because we may choose $n$ big enough. In the case, $n$ is fixed, then the conjecture may be false. For example, at $n=2,q=5$, there is no such irreducible polynomial. What do you think?
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$X^p-X-1$ works if $q = p^{f}$ and $f$ is not divisible by $p$, see https://en.wikipedia.org/wiki/Artin%E2%80%93Schreier_theory – Jef Jan 25 '18 at 23:39
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Thanks, I know that result also. I am looking for the answer for the case f is divisible by p as well. In this case, I believe the existence of irreducible polynomial of the form $x^n-x-1$ for n suitable (but not for all $n$: Calculations on Maple showed me that $n\neq 10$ if $p=3$). Besides, the polynomial $f(x)=x^p-x-1$ has exponent (order) $\frac{p^p-1}{p-1}$, Hence, we may construct some polynomials with bigger degree of the form $f(x^t)$ with suitable odd numbers $t$ by appling the following lemma: – Hieu Jan 26 '18 at 13:39
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LEMMA: $$f(x^t) \text{ is irreducible iff } \begin{array}{ll}1. gcd(t,\frac{p^p-1}{e})=1 \text{ and } \2. p \text{ prime, } p\mid t \Rightarrow p\mid e \text{ and }\ 3. 4 \mid t \Rightarrow 4 \mid (p^2-1)\end{array} $$ – Hieu Jan 26 '18 at 13:42