1.Introduction

It has been proven that all integers except 9n+4 and 9n-4 are representable by 
the sums of four cubes. ([1]. Demjanenko. )
Well, are all integers except 9n+4 and 9n-4 representable by the sums of three cubes?
The following integers cannot be solved within the range of d<1000 yet. 
d=33, 42, 74, 114, 156, 165, 318, 366, 390, 420, 564, 579, 627, 633, 732, 758, 789,
795, 894, 906, 921, 933, 948, 975 

On the other hand, many results are known within the range of d<1000. 
([2]. Bernstein. )

We can easily show that the integer of 9n+4 and 9n-4 are not representable
by the sums of three cubes. 
Let x,y,and z are integer,then x3=0,1, and -1 mod 9.
It is also similar y3 and z3.
So, x3+y3+z3 not =4,-4 mod 9. 


[1].Demjanenko:http://www.math.u-bordeaux1.fr/~cohen/sum4cub.ps 
[2].Bernstein:http://cr.yp.to/threecubes.html           
       
2.Method

Solving of x3+y3=z3+d  is possible to return in the problem of solving z3+d=0 mod n
in addition as x+y=n. 
To solve for z3+d=0 mod n, we need to obtain phi(n) and solve for 3u-phi(n)v=1.
So,we can obtain z,though it should be (3,phi(n)) =1 of course.
We define phi(n) as euler's totient function of n.

x3+y3=z3+d is able to be solved if there is integer solution of simultaneous equation
{x+y=n,x2-xy+y2=(z3+d)/n)}. 

I found the easy formulae that gave the solution of x3=m mod p as shown below.


     Formulae for the solution of x3=m mod p

     x,m: integer
     p  : prime
     (p,m)=1　　
     

      Case (1).  p=2 mod 3

                x=m(2p-1)/3 mod p. 

     Case (2).  p=4 mod 9 and if  m(p-1)/3=1 mod p 

                x=m(2p+1)/9 mod p, and two roots of the remainder = xw,xw2  mod p. 
                w is assumed to be the root of x2+x+1=0 mod p. 

     Case (3).  p=7 mod 9 and if  m(p-1)/3=1 mod p 

                x=m(p+2)/9 mod p, two roots of the remainder = xw,xw2  mod p. 

Proof.

     Case (1). Let x=m(2p-1)/3

　       　　x3=（m(2p-1)/3)3
　　　         = m(2p-1)
　　　         = mpm(p-1)
　　　         = m　mod　p　(Fermat's little theorem：m(p-1)=1 mod p)


     Case (2). Let x=m(2p+1)/9

　　　      x3=（m(2p+1)/9)3
　　　　　　　= m(2p+1)/3
　　　　　　　= m（m(p-1)/3)2
　　　　　　　= m　mod　p
　　
    Case (3). Let x=m(p+2)/9

            x3=（m(p+2)/9)3
            　= m(p+2)/3
　　　　　  　= mm(p-1)/3
　　　　　　  = m　mod　p


    Q.E.D.



 

3. Example


Example 1. Find the solutins of　x3=2 mod 31===> x={4,7,20}

　　　     p=4 mod 9,then
　　　     x=2(2*31+1)/9 mod 31
　　　　  　=4　mod 31

　　     　solutions of x2+x+1=0 mod 31 are {5,25},then
 
　　       other solutions are 4*5=20,4*25=7 mod 31,then

　　           　x={4,7,20} mod 31

Example 2. Find the solutin of  x3+y3-z3=52===>  607029013173 + 239612924543 -619227128653=52

　　　Suppose n=x+y=60702901317 + 23961292454=84664193771=521*162503251 
　 
     If we solve z3+52=0 mod n,but n is composite so we must find the prime factor of n.
     We find n=521*162503251
　　
     Soulution of z3+52=0 mod 521 is, z=(-52)((2*521-1)/3) mod 521,
     then   z=290 mod 521

     Soulutions of z3+52=0 mod 162503251 are, z=(-52)((162503251+2)/9) mod 162503251,
     then z={8974234,86694519,66834498} mod 162503251
      

     By  CRT(Chinese Remainder Theorem),solutions of z3+52=0 mod n are
　　 z={ 25417341654, 37462442249, 61922712865}
     At z=61922712865,solve the equation{x+y=84664193771,x2-xy+y2=(619227128653+52)/84664193771)}
     then  x=60702901317,y=23961292454

     We obtained {x,y,z}={60702901317,23961292454,61922712865} as the solution of　x3+y3-z3=52.