(I) Three nonzero integers x<y<z are in Arithmetic Sequence (in this order).
Determine all possible triplets (x,y,z) that satisfy this equation;
x
2 + y
2 = z
2 - 2x*y*z
(II) With all the other conditions in (I)remaining unaltered, determine all possible triplets that satisfy this equation:
x3+z3=y3+7*y*z
Equation (I) Let x=y-d and z=y+d. Requiring d>0 will have x<y<z in accordance to the problem statement.
Then x^2 + y^2 = z^2 - 2xyz becomes (y-d)^2 + y^2 = (y+d)^2 - 2(y-d)y(y+d).
This reduces to y^2 + 2y^3 = 4yd + 4yd^2. Since y is nonzero, we can divide by y: y + 2y^2 = 4d + 4d^2.
Multiply both sides by 8 and complete the squares: (4y+1)^2 + 15 = (4d+4)^2
So at this point we need squares that differ by 15. There are two options: 1^2 + 15 = 4^2 and 7^2 + 15 = 8^2.
Case 1 has +/-1 = 4y+1, which makes y=0 or -1/2, both of which must be discarded.
Case 2 has +/-7 = 4y+1, which makes y=-2 or 3/2. 3/2 must be discarded but y=-2 is a valid part of a solution. Then +/-8 = 4d+4 makes d=1 or -3. -3 needs to be discarded leaving d=1.
Then with y=-2 and d=1 we have one solution fulfilling all the conditions: (x,y,z) = (-3,-2,-1).
Equation (II) Same as before, let x=y-d and z=y+d with d>0.
Then x^3 + z^3 = y^3 + 7yz becomes (y-d)^3 + (y+d)^3 = y^3 + 7y(y+d).
This reduces to y^3 - 7y^2 + 6yd^2 - 7yd = 0. Since y is nonzero, we can divide by y: y^2 - 7y + 6d^2 - 7d = 0.
Multiply both sides by 24 and complete the squares: 6*(2y-7)^2 + (12d-7)^2 = 343
So at this point we need squares so that six times the first plus the second equal 343. There are two options 6*3^2 + 17^2 = 343 and 6*7^2 + 7^2 = 343.
Case 1 has +/-3 = 2y-7, which makes y= 2 or 5. Then +/-17 = 12d-7 makes d=2 or -5/6. d=-5/6 must be discarded. That leaves two potential solutions y=2 or y=5 both with d=2.
The first value of y gives us (x,y,z)=(0,2,4) but must be discarded as having x=0. But the second choice of y gives us a solution fulfilling all the conditions: (x,y,z) = (3,5,7).
Case 2 has +/-7 = 12d-7, which makes d=0 or 7/6. Both values must be discarded so this case yields no new solutions.
Edited for formatting.
Edited on March 29, 2023, 12:32 am
Analytic Solution
