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%I #3 Mar 31 2012 14:40:24
%S 19,4513,14689,32401,26929,48019,44641,72739,124099,179683,211249,
%T 288979,395089,386131,587233,905059,1040419,1410049,2237011,1919779,
%U 2078209,2220451,2950963,2767489,4919971,5582449,5019889,5255761
%N Let a,b be prime numbers satisfying the Diophantine equation a^3+b^3=(a+b)*(a^2-a*b+b^2)=c^2. Then the second factor a^2-a*b+b^2 is 3*e^2 for some integer e. This sequence tabulates the 'e' values, sorted by magnitude of c.
%C For each n let a=A099806[n], b=A099807[n], c/12=A098970. Then a^3+b^3=c^2. The left side factors as (a+b)*(a^2-a*b+b^2). The second factor is 3*e^2 for some integer e. The sequence tabluates the 'e' values. These 'e' values all have the form 3*M^4+N^4, for some pair M,N of relatively prime integers of opposite parity. Remember, a and b are prime numbers.
%H James Buddenhagen, <a href="http://www.buddenbooks.com/jb/num_theory/sum_of_2_cubes_a_square.htm">Two Primes Cubed which Sum to a Square</a>.
%e 11^3+37^3=228^2, 11^2-11*37+37^2=3*e^2 with e=19, so 19 is in the sequence.
%Y Cf. A099806, A099807, A098970, A099808.
%K nonn
%O 0,1
%A _James R. Buddenhagen_, Oct 26 2004