Find all possible positive integers A, B and C - with A ≤ B, that simultaneously satisfy:

- A+B-C=12, and:
- A
^{2} + B^{2} - C^{2} = 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)**

**/*************************/**

**OOPs. I got the signs wrong on equation (1). Revised solution coming up.**

*Edited on ***November 1, 2015, 4:05 pm**