Find all possible positive integers A, B and C - with A ≤ B, that simultaneously satisfy:
- A+B-C=12, and:
- A2 + B2 - C2 = 12
(1) C = A+B-12
(2) Substituting in the 2nd equation and simplifying gives
-156 = 24A + 24B - 2AB
(3) Solving for B gives B = (-78-12A)/(12-A)
(4) Let D = 12-A. Then B = 12 - 66/D
(5) The only possible values for D are +/- 1,2,3,6,11,22,33,66.
So there are 16 possible solutions.
They are:
D B A C
1 -54 11 -55
2 -21 10 -23
3 -10 9 -13
6 1 6 -5
11 6 1 -5
22 9 -10 -13
33 10 -21 -23
66 11 -54 -55
-1 78 13 79
-2 45 14 47
-3 34 15 37
-6 23 18 29
-11 18 23 29
-22 15 34 37
-33 14 45 47
-66 13 78 79
The only solutions with A, B and C positive and A <= B are
(13,78,79), (14,45,47), (15,34,37) and (18,23,29)
Corrected Analytical Solution
