[seqfan] Solvability of equation x^3+y^3+z^3=0 in integers of real and imaginary quadratic fields

[Vladimir Shevelev]

Dear SeqFans,

I just have submitted two, in my opinion, interesting sequences:

%I A180323
%S A180323 2,5,6,11,15,17,29,33,41,42,43,51,53,58,62,65,69,82,85,86,89,93
%N A180323 Positive numbers l of the form l=A007913(k^4-4*k*m^3), where 1<=k<=5*l, 1<=|m|<=5*l. 
%C A180323 Equation x^3+y^3+z^3=0 is solvable in numbers of the form N+M*sqrt(a(n)), where M and N are integers. Moreover, it is solvable in numbers of the form N+M*sqrt(l), where l has the form l=A007913(k^4-4*k*m^3), where k,|m|>=1 (without restictions k,|m|<=5*l). But in this more general case could be unknown numbers l having this form; this circumstance does not allow to construct full sequence of such l. Therefore, we restrict ourself by condition k,|m|<=5*l. Note that testing l with respect to this condition is rather simple by sorting all values of k,|m|<=l. One can prove that, at least, in case that Fermat numbers (A000215) are square free, then the sequence is infinite. Conjecture(necessity of the form of l). If equation x^3+y^3+z^3=0 is solvable in numbers of the form N+M*sqrt(l) with integer N,M, then there exist positive integers k,m such that l=A007913(k^4-4*k*m^3). 
%F A180323 Let a(n)=A007913(k^4-4*k*m^3). Put g=sqrt(A008833(k^4-4*k*m^3)). Then identity A^3+B^3+C^3=0 is valid, where A=2*m^6-k^3*m^3-k^6+k*(k^3+5*m^3)*g*sqrt(a(n)); B=3*m*(k^3-m^3)*(k^2-g*sqrt(a(n))); C=k^6-8*k^3*m^3-2*m^6-k*(k^3-4*m^3)*g*sqrt(a(n)). 
%Y A180323 A007913 A008833 A000215 
%K A180323 nonn
%O A180323 1,1

 
%I A180336
%S A180336 2,3,5,6,11,15,23,31,47,51,59,86
%N A180336 Positive numbers l of the form l=A007913(4*k*m^3-k^4), where 1<=k<=5*l, 1<=|m|<=5*l. 
%C A180336 Equation x^3+y^3+z^3=0 is solvable in numbers of the form N+M*sqrt(-a(n)), where M and N are integers. Moreover, it is solvable in numbers of the form N+M*sqrt(-l), where l>0 has the form l=A007913(4*k*m^3-k^4), where k,|m|>=1 (without restictions k,|m|<=5*l). But in this more general case could be unknown numbers l having this form; this circumstance does not allow to construct full sequence of such l. Therefore, we restrict ourself by condition k,|m|<=5*l. Note that testing l with respect to this condition is rather simple by sorting all values of k,|m|<=l. One can prove that, at least, in case that Fermat numbers (A000215) are square free, then the sequence is infinite. Conjecture(necessity of the form of l). If equation x^3+y^3+z^3=0 is solvable in numbers of the form N+M*sqrt(-l) with integer N,M, then there exist positive integers k,m such that l=A007913(4*k*m^3-k^4). 
%F A180336 Let a(n)=A007913(4*k*m^3-k^4). Put h=sqrt(A008833(4*k*m^3-k^4)). Then identity A^3+B^3+C^3=0 is valid, where A=2*m^6-k^3*m^3-k^6+k*(k^3+5*m^3)*h*sqrt(-a(n)); B=3*m*(k^3-m^3)*(k^2-h*sqrt(-a(n))); C=k^6-8*k^3*m^3-2*m^6-k*(k^3-4*m^3)*h*sqrt(-a(n)). 
%e A180336 We have 2=A007913(4*k*m^3-k^4) for k=2,m=3. Therefore, a(1)=2; furthermore, 3=A007913(4*k*m^3-k^4) for k=m=1. Therefore, a(2)=3. 
%Y A180336 A180323 A007913 A008833 A000215 
%K A180336 nonn
%O A180336 1,1

More terms? References? (I did all results myself) Corrections?

Regards,
Vladimir

 Shevelev Vladimir‎