(x,y,z) = (1, 2, 3) and (7, 8, 9) constitutes the only possible solution to the given equation.
EXPLANATION:
By the conditions of the problem:
z = y + a and x = y –a , where a is any positive integer less than x.
Accordingly, in terms of provisions of the problem under reference:
(y + a)^4  64 = (y  a)^ 4 + y^4
or, 8*(y^3)* a + 8*(a^3)*y = y^4 + 64(#)
It is clearly observed from (#) that y is even ,so that substituting y = 2*p , we obtain :
4*(p^3)*a + (a^3)*p  4 = p^4
Or, p(4*(p^2)*a + a^3  p^3) = 4; so that p must be a positive divisor of 4 giving p= 1,2,4
Now; p=1 gives, a^3 + 4*a  5 = 0, which yields: (a1)((a^2)+ a + 5) =0; giving a = 1, (1 +/ sqrt(19))/2, so that the only integral root in this case is given by a = 1 Hence, y=2*p = 2 Consequently, z= y+1 = 3 and x = y 1 = 1
p=2 gives, (a^3) + 16*a  10 = 0, which does not possess any positive integer solution in a.
Now; p=4 gives, 4*(a^3) + 256*a  4 = 256
or, (a^3) + 64*a  65 = 0, giving:
(a1)*((a^2) +a + 65)= 0, giving a=1 or, a= (1+/ sqrt(16899))/2.
Since, a is a positive integer, it follows that a=1; giving:
y=2*p = 2*4 = 8.
Consequently, z= y+1 = 9 and x = y 1 = 7
Therefore, (x, y, z) =(1, 2, 3) and (7, 8, 9) constitutes the only positive integral solutions to the equation under reference.
Also refer to Bractals' solution and explanation.
