|
%I #4 Mar 31 2012 21:03:56
%S 101,127,137,307,379,487,571,617,643,701,761,859,881,1013,1039,1217,
%T 1229,1231,1277,1361,1447,1831,2081,2179,2239,2417,2467,2477,2621,
%U 2861,2971,3257,3413,3449,3461,3559,3583,3701,3907,4013,4049,4133,4219,4241
%N Primes p such that f_p(x)=(1296+432*x+108*x^2+24*x^3+5*x^4+x^5) mod p factors as product of 3 linear and one irreducible quadratic factor.
%C The fact that f_101 factors as a product of 3 linear and one irreducible quadratic factor shows that the Galois group of f(x) is (isomorphic to) the Symmetric group on 5 letters, S_5. That is also the Galois group of 1+2*x+3*x^2+4*x^3+5*x^4+6*x^5.
%D N. Jacobson, Basic Algebra I, Freeman and Co, (1985), pp. 301-304.
%e When p=101, f_p(x)=(x+40)*(x+30)*(x+49)*(x^2+88*x+72) mod p
%K easy,nonn
%O 1,1
%A _Waldeck Schutzer_, Oct 13 2002
|
