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Jul 11, 2015 at 18:28 comment added Dracula x^i + x - 1 with i even is irreducible mod 3 for i = 2,4,6,14,30,54,364,446,638,1382,1478,2726, and ??.
Jul 11, 2015 at 8:34 comment added Ofir Gorodetsky Mine neither. But I think that the obstruction for such polynomials to be irreducible comes from certain possible quadratic factors only (proposition 3.1.2). If you can rule out those factors (for infinitely many $(a,b)$'s), it establishes irreducibility. It also might be of use to read other papers by Agou, or to contact him.
Jul 11, 2015 at 8:28 comment added Pablo Do you understand their conclusion? Sadly enough, my French is not perfect yet...
Jul 11, 2015 at 8:21 comment added Ofir Gorodetsky I see :) Then I suggest trying to find a suitable family. What about $a\cdot X^{3^k+1}+X+b$? The following paper of Agou, in French, studies the irreducibility of that family: matwbn.icm.edu.pl/ksiazki/aa/aa44/aa4444.pdf (MR number is MR0777011). If you delve into this, I would be glad to see an update later!
Jul 11, 2015 at 8:13 comment added Pablo Ofir, thank you very much for all this information! The point is that I have already thought about some of this things, and came to the conclusion that treating this problem as restrictions on coefficients is not the right way. The reason for this is that it seems not so difficult to prove that there are infinitely many irreducible polynomials of the form $f(X^2)$ directly, but doing this with the methods of saying "all the odd coefficients = 0" is probably hopeless. I do agree that taking half of the coefficients to be zero and hoping for an irreducible polynomial is too much to ask for.
Jul 11, 2015 at 7:58 history edited Ofir Gorodetsky CC BY-SA 3.0
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Jul 11, 2015 at 7:53 history edited Ofir Gorodetsky CC BY-SA 3.0
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Jul 11, 2015 at 7:44 history edited Ofir Gorodetsky CC BY-SA 3.0
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Jul 11, 2015 at 7:39 history edited Ofir Gorodetsky CC BY-SA 3.0
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Jul 11, 2015 at 7:33 history edited Ofir Gorodetsky CC BY-SA 3.0
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Jul 11, 2015 at 7:25 history answered Ofir Gorodetsky CC BY-SA 3.0