I) Substituting z = y+a and x = y-a, we obtain:
y(4a-y) = 135
Or, 4y = 135/y + y...(i)
Since, x, y and z are positive integers, it follows that y > a.
So, 4y > 4a = y + 135/y; giving:
y^2 > 45 so that y > 6.
Divisors of 135 greater than 6 are 9, 15, 27,45 and 135.
Now, y = (9, 15, 27, 45, 135) gives a=(6, 6, 8, 12, 34) from (i).
So, (x,y,z) = (3,9,15); (9,15,21); (19,27,35); (33,45,57); (101,135,169)
II) Substituting z = y+a and x = y-a, we obtain:
y(4a-3y) = 105
Or, 4y = 105/y +3 y...(ii)
Since, x, y and z are positive integers, it follows that y > a.
So, 4y > 4a = 3y + 105/y; giving:
y^2 > 105 so that y > 10.
Divisors of 105 greater than 10 are 15, 21,35 and 105.
Now, y = ( 15, 21, 35, 105) gives a=(13, 17, 27, 79) from (ii).
So, (x,y,z) = (2, 15, 28); (4, 21, 38); (8, 35, 62); (26, 105, 184)
III) Substituting z = y+a and x = y-a, we obtain:
y(4a-y) = y(y^2 - a^2)
Or, a^2+4a = y^2+y (since y is a positive integer, y cannot be 0)
Or, A^2 - B^2 = 15 where A = 2a+4 and B = 2y+1......(iii)
Now, (iii) possess precisely two solutions in positive integers which are (A,B) = (4,1); (8,7)
But, (A,B) = (4,1) gives y =0 which is a contradiction.
(A,B) = (8,7) gives a = 2 and y=3, so that:
(x,y,z) = (1,3,5) which is the only possible solution to the given equation.
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