On April 27, 2008,I found that 4+27=31=prime,22+33=31=prime.

I searched the solutions of y2+x3 = p by algebraic method because generally it was often
being solved by using the elliptic curve.

I got the following result when I looked for other solutions of y2+x3 = p.

About some p which doesn't have the solution,we can prove that such p
does not have a integer solution by the next theorem([1]. Alaca and Williams).


Theorem 1.

condition
       1. k>0
       2. k is squarefree
       3. k=2,3 mod 4
       4. class number of  Q(sqrt(k)) <> 0 mod 3
       5. T+Usqrt(k) is fundamental unit of  Q(sqrt(k)) of norm 1

 IF k=4 mod 9,U=0 mod 9 or k=7 mod 9,U=+-3 mod 9 or k=4 mod 7,U=0 mod 7 
      then the equation y2=x3 + k has no solutions in integer x and y.



    When we apply the above theorem to y2+x3 = p, we get next results.

    Case1. k=4 mod 9,U=0 mod 9

           p=67,103,139,571,607,751,and 787 have no integer solution.

    Case2. k=7 mod 9,U=+-3 mod 9

           p=7,367,439,547,619,691,and 907 have no integer solution.

    Case3. k=4 mod 7,U=0 mod 7

           p=683,823,and 991 have no integer solution.


Theorem 2.

condition
       1. m=3 mod 4
       2. n=2 mod 4
       3. k=m3-n2
       4. prime factor of n/2 = 1 mod 4


    Then the equation y2=x3 + k has no solutions in integer x and y.

       p=11 and 23 have no integer solution.


Theorem 3.

condition
       1. m=2 mod 4
       2. n=1 mod 2
       3. k=m3-n2
       4. prime factor of n = 1 mod 4


    Then the equation y2=x3 + k has no solutions in integer x and y.

       p=7,47,191 and 503 have no integer solution.

     
{1].Alaca and Williams: Introductory Algebraic Number Theory





                       Integer solutions  of  y2+x3 = p 

 
                                 
                                 p: prime
                                 p < 1000 

                                 abs(x) < 1000000 
                                 T,U: Fundamental solution of T2-pU2=1