A099808 - OEIS

A099808

If a,b are primes which satisfy the Diophantine equation a^3 + b^3 = c^2, then this sequence consists of the numbers sqrt((a+b)/48), sorted by the magnitude of c.

%I #5 Mar 31 2012 14:40:24

%S 1,15,28,35,44,44,55,56,91,90,88,119,161,165,200,184,273,319,285,357,

%T 377,400,380,434,550,517,592,615,638,667,682,666,740,697,784,688,825,

%U 682,846,770,893,814,868,925,775,899,885,1007,1045,1040,1078,1184,1015

%N If a,b are primes which satisfy the Diophantine equation a^3 + b^3 = c^2, then this sequence consists of the numbers sqrt((a+b)/48), sorted by the magnitude of c.

%C For each n let a=A099806[n], b=A099807[n]. Then sqrt((a+b)/48) is an integer and equals A099808[n]. Note that a^3 + b^3 = c^2 factors as (a+b)*(a^2-a*b+b^2). The first factor (a+b) is 48*d^2, some d. This sequence tabulates the d values. 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 From 11^3 + 37^3 = 228^2 we get sqrt((a+b)/48)=(11+37)/48=1, so 1 is in the sequence.

%Y Cf. A099806, A099807, A098970, A099809.

%K nonn

%O 0,2

%A _James R. Buddenhagen_, Oct 26 2004

%E Example corrected by Harvey P. Dale, Apr 12 2011.