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I came up with this problem and have found it interesting.

Problem. For what integers $m\gt n\gt 0$, the polynomial $f(x)=x^m+x^n+1$ is irreducible in $\mathbb Q[x]$?

If $mn\equiv 2 \pmod 3$, i.e. one of $m,n $ is congruent to $1 \bmod3$ and the other is $2 \bmod 3$, then $f(x)$ is divisible by $x^2+x+1$, hence reducible (unless $m=2$).
Moreover, let $d$ be the largest power of 3 to divide $m$ and $n$, $m_0=m/d$ and $n_0=n/d$, if $m_0n_0\equiv 2 \pmod 3$ and the condition "$\exists k, m=2n=2\times 3^k$" fails, then $f(x)$ is reducible.
But there are many other cases to consider, and I don't know what to do. Thanks for any help in advance.

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I will reproduce here my answer [copied from Math Reviews] to https://mathoverflow.net/questions/56579/about-irreducible-trinomials (one should also see the other answer posted to that question, as well as a useful comment):

MR0124313 (23 #A1627) Ljunggren, Wilhelm On the irreducibility of certain trinomials and quadrinomials. Math. Scand. 8 1960 65–70. 12.30

The author considers the irreducibility over the field of rational numbers of the polynomials $f(x)=x^n+ε_1x^m+ε_2x^p+ε_3$, where $ε_1,ε_2,ε_3$ take the values $\pm1$. He proves that if $f(x)$ has no zeros which are roots of unity, then $f(x)$ is irreducible; if $f(x)$ has exactly $q$ such zeros, then $f(x)$ can be factored into two factors with rational coefficients, one of which is of degree $q$ with all these roots of unity as zeros, while the other is irreducible (and possibly merely a constant). He also determines all possible cases where roots of unity can be zeros of $f(x)$. As a corollary he is able to give a complete treatment of the trinomial $g(x)=x^n+ε_1x^m+ε_2$, where $ε_1,ε_2$ take the values $\pm1$. The irreducibility of this trinomial was studied by E. S. Selmer, who gave a partial solution [Math. Scand. 4 (1956), 287--302; MR0085223 (19,7f); see also #A1628]. The methods used are direct and elementary. Reviewed by H. W. Brinkmann

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