Integer solution?

Look in mod 7...........

The numbers $a,a+1,a+2,a+3,a+4,a+5,$ and $a+6$ make a complete set of residues modulo $7$ . Therefore, when we add up their cubes, it is the same as adding $0^{3}+1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}\equiv 0\pmod{7}$ Now consider the sum of two consecutive fourth powers modulo $7$ . The fourth powers modulo $7$ are $0,1,2,4,4,2,$ and $1$ . There are no two consecutive fourth powers that add to $0\pmod{7}$ , so there are no solutions.