A223942 Least prime q such that (x^{p_n}-1)/(x-1) is irreducible modulo q, where p_n is the n-th prime.

2, 2, 2, 3, 2, 2, 3, 2, 5, 2, 3, 2, 7, 3, 5, 2, 2, 2, 2, 7, 5, 3, 2, 3, 5, 2, 5, 2, 11, 3, 3, 2, 3, 2, 2, 7, 5, 2, 5, 2, 2, 2, 19, 5, 2, 3, 2, 3, 2, 7, 3, 7, 7, 11, 3, 5, 2, 43, 5, 3

Offset: 1

Author

Zhi-Wei Sun, Mar 29 2013

Keywords

• nonn

Comments

It is well known that (x^{p^n}-1)/(x^{p^{n-1}}-1) is irreducible over the rationals for any prime p and positive integer n.

We have the following "Reciprocity Law": For any positive integer n and primes p > 2 and q, the cyclotomic polynomial (x^{p^n}-1)/(x^{p^{n-1}}-1) is irreducible modulo q if and only if q is a primitive root modulo p^n.

This can be proved as follows: As any monic irreducible polynomial over F_q=Z/qZ of degree k divides x^{q^k}-x in the ring F_q[x], the polynomial f(x)= (x^{p^n}-1)/(x^{p^{n-1}}-1) in F_q[x] has an irreducible factor of degree k < deg f if and only if f(x) is not coprime to x^{q^k}-x for some k < p^n-p^{n-1}. Note that gcd(x^{p^n}-1,x^{q^k-1}-1) = x^{gcd(p^n,q^k-1)}-1. If p^n | q^k-1, then x^{p^n}-1 | x^{q^k}-x and hence f(x) divides x^{q^k}-x; if p^n does not divide q^k-1, then gcd(x^{p^n}-1,x^{q^k-1}-1) divides x^{p^{n-1}}-1 and hence f(x) is coprime to x^{q^k}-x. Thus, f(x) is irreducible modulo q, if and only if p^n | q^k-1 for no 0 < k < p^n-p^{n-1}, i.e., q is a primitive root modulo p^n.

By the above "Reciprocity Law" in the case n=1, we have a(k) = A002233(k) for all k > 1.

Conjecture: a(n) <= sqrt(7*p_n) for all n>0.

Examples

			  a(9)=5 since f(x)=(x^{23}-1)/(x-1) is irreducible modulo 5, but reducible modulo either of 2 and 3, for,
   f(x)==(x^{11}+x^9+x^7+x^6+x^5+x+1)
         *(x^{11}+x^{10}+x^6+x^5+x^4+x^2+1) (mod 2)
and
   f(x)==(x^{11}-x^8-x^6+x^4+x^3-x^2-x-1)
         *(x^{11}+x^{10}+x^9-x^8-x^7+x^5+x^3-1) (mod 3).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..450

Crossrefs

Cf. A000040, A002233, A122028, A217785, A217788, A218465, A220072, A223934.

Programs

Mathematica

Do[Do[If[IrreduciblePolynomialQ[Sum[x^k,{k,0,Prime[n]-1}],Modulus->Prime[k]]==True,Print[n," ",Prime[k]];Goto[aa]],{k,1,PrimePi[Sqrt[7*Prime[n]]]}];
Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]