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 = 12-A-B
(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
-66 13 78 -79
-33 14 45 -47
-22 15 34 -37
-11 18 23 -29
-6 23 18 -29
-3 34 15 -37
-2 45 14 -47
-1 78 13 -79
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
The only solutions with A, B and C positive are
(1,6,5) and (1,5,6)
The only solution with A <= B is (1,6,5)
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OOPs. I got the signs wrong on equation (1). Revised solution coming up.
Edited on November 1, 2015, 4:05 pm
Analytical solution (spoiler)
