(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:
(a-1)((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:
(a-1)*((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.
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