Solution of (I):
Since x, y and z are positive integers in arithmetic sequence with z>y>x, we can substitute z = y+a and x = y-a where a is a positive integer.
So, x^3+z^3 = y^3 + 10yz
Or, 2y(y^2 + 3*a^2) = y^3 + 10y(y+a)
Or, y^2 + 6*a^2 = 10(y+a)( since y is positive, y=0 ia a contradiction)
Or, 6*A^2 + B^2 = 175, where (A,B) = (y-5, 6a-5)
Or, (A,B) = (1,13),(-1,13), (1,-13), (-1,-13), (3,11), (-3,11), (3,-11), (-3,-11), (5,5), (-5,5), (5,-5),(-5,-5)
Of the abovementioned choices of (A,B), only (A,B) = (1,13), (-1,13) yields positive integer values for y and a.
Now, (y-5, 6a-5) = (1,13); (-1,13) gives:
(y,a) =(6,3), (4,3), so that:
(x,y,z) = (3,6,9), (1,4,7) constitutes all possibl;e solutions to the given problem.
Solution of (II):
Let x = y-a and z = y+a
Then,x^3 + y^3 = z^3 - 2 yields:
y^2(y-6a) = 2(a^3-1)...(i)
Since, LHS must be divisible by 2, it follows that y is even. So substituting y = 2p, we obtain:
4*p^2(p-3a)+ 1 = a^3
Or, a^3 = 1 (Mod 4)
Hence, y is even and a is odd....(ii)
Now, if y<6a then, LHS of (i) is negative, while RHS of (i) is non-negative. This is a contradiction.
Hence, y>=6a.
Case A: y = 6a
From (i), we obtain:
a^3 = 1, giving a=1, so that y=6, yielding:
(x,y,z) = (5,6,7)
Case B: y is greater than 6a
Minimum value of y is (6a+1), so that:
(6a+1)^2 <= y^2(y-6a)= 2(a^3-1)
So, 2*a^3 >= 36*a^2 + 12a +3 > 36*a^2
Or, a^3> 18*a^2, giving:
a>18, so that a>=19, yielding:
y> 6a+1>= 115, so that y>=116 as y must be even
Consequently, (x,y,z) = (5,6,7) is the only possible solution whenever y<116.
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