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(A, B, C) = (1,4,5), (1,5,4), (4,5,1), (4,1,5), (5,1,4), (5,4,1)(1,3,8), (1A,8,3), (3,8,1), (3,1,8), (8,1,3), (8,3,1), (2,2,3), (-4,1,1), (1,-4,1), (1,1,-4) are all possible solutions corresponding to the given problem.
EXPLANATION
At the outset, consider that each of A, B, and C is a negative integer.
Then, (A, B, C) = (-P, -Q, -R), where each of P, Q, and R is a positive integer.
Then, the given equation reduces to:
(1-1/P)(1-1/Q)(1-1/R) = 3
Since each of P, Q, and R is a positive integer, it follows that: each of (1-1/P), (1-1/Q), (1-1/R) is <1, so that: (1-1/P)(1-1/Q)(1-1/R) <1<3.
This is a contradiction.
Therefore, we will consider the cases where:
(A) Each of A, B, and C is a positive integer
(B) P, Q, and R is a combination of positive and negative integers.
CASE (A):
Each of the variables A, B, and C is a positive integer.
We assume that each of A, B, and C is >=3
Then, we must have:
(1+1/A)(1+1/B)(1+1/C) <= (4/4).(4/3).(4/3) = 64/27<3.
This is a contradiction.
Thus, at least one of A, B, and C must be <3
We now impose the restriction A<= B <=C
(i) Let A =1
Then, from the given equation, we must have:
(1+1/B)(1+1/C) = 3/2
=> 1+1/B = (3C)/{2(C+1)}
=> 1/B = (C-2)/(2C+2)
=> B = (2C+2)/(C-2) = 2 + 6/(C-2) ..........(*)
So, C-2 must divide 6.
Thus, C-2 = 1,2 3, 6
--> C= 3,4,5,8
Now, from(*), we have:
C=3 gives: B = 8, which violates the restriction B<=C
C= 4 gives: B = 5, which violates B<= C
C=5 gives: B = 4
C = 8 gives: B= 3
Then we have (A, B, C) = (1,4,5), (1,3,8) with the restriction A<=B <= C
(ii) Let A =2
--> (1/1/A) = 3/2, so that:
(1+1/B)(1+1/C) = 2
=> 1+ 1/B = (2C)/(C+1)
=> B = (C+1)/(C-1)= 1+ 2/(C-1) ........(**)
So, C-1 must divide 2
=> C-1 = 1,2
=> C = 2,3
Now, from (**), we have:
C=2 gives: B= 3 which violates B<= C
C=3 gives B=2
Then (A, B, C) = (2,2,3) (iii) If possible, let A >2
=> (1+1/A) > 3/2
Then, (1+1/A)(1+1/B)(1+1/C) =3 gives:
( 1+1/B)(1+1/C) < 2*3/3 = 2.......(***)
Since each of B and C must be >= A, we must have each of B and C must be >2
In that situation, we have:
(1+1/B)(1+1/C) > (3/2).(3/2) = 9/4 >2. This violates (***).
Accordingly, this leads to a contradiction.
Then, with the restriction A<=B<=C, the solutions are:
(A, B, C) = (1,4,5), (1,3,8), and (2,2,3)
So, the unrestricted solutions are all possible permutations of the above and are given by:
So, the unrestricted solutions are all possible permutations of the above and are given by:
(A, B, C) = (1,4,5), (1,5,4), (4,5,1), (4,1,5), (5,1,4), (5,4,1),(1,3,8), (1,83,), (3,8,1), (3,1,8), (81,3), (8,3,1)
the restriction A<= B <= C
CASE -(B)
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Each of the variables is a combination positive integers and negative integers.
(i) Two of the variables are negative and the remaining is positive.
Since the given equation is symmetric, we can WOLG assume that each of A and B is a negative integer and C is a positive integer
Then, (A, B) = (-P, -Q), where each of P and Q is a positive integer.
Hence, (1-1/P)((1-1/Q)((1+1/C) =3
Now, C is a positive integer. So, 1/C <=1, so that:
(1+1/C) <=2
Then, we must have:
(1-1/P)(1-1/Q) >=3/2 ........(***)
However, each of (1-1/P) and (1-1/Q) must be <1, so that (1-1/P)(1-1/Q) is <1 <3. This violates (***).
Therefore, this sub-case does NOT have any valid solutions.
(iii) Two of the variables is a positive integer and the remaining one is a negative integer.
Since, the equation is symmetric, we can WLOG consider A to be negative.
Then, A = -P, where P is a positive integer.
Since each of B and C is a positive integer, it follows that each of 1/B and 1/C is <=1, so that:
(1+1/B)(1+1/C) >=4
Accordingly, from the given equation, we must have:
(1-1/P) >=3/4
=> P <=4 ..........(+)
If P=4, then: (1+1/B)(1+1/C) =4
=> 1+1/B = (4C)/(C+1)
=> 1/B = (3C-1)/(C+1)
=> B = (C+1)/(3C-1)
Since B is a positive integer, then:
C+1 >= 3C-1
=> 2C <= 2 => C <= 1
However, C is a positive integer, so that C >= 1
Thus, the two inequalities are simultaneously satisfied whenever C=1, giving B =2/2 =1
Thus, (A, B, C) = (-4,1,1) is a solution.
Also, as C is a positive integer, it follows that:
C=>1. Also, each of P and Q is a positive integer, that is, each of P and Q is >=1, so that:
(1+1/P)(1+1/Q)<=4
If P=3, then (1-1/P) = 2/3
Then, ((1-1/P)(1+1/B)(1+1/C) = 3
gives: (1+1/B)(1+1/C) = 3.3/2 =9/2>4, which leads to a contradiction.
If P=2, then (1-1/P)= 1/2
This gives: (1+1/B)(1+1/C) = 6 >4, which leads to a contradiction.
If P=1, then (1-1/P) = 0, so that:
(1-1/P)(1+1/B)(1+1/C) = 0 <3., which leads to a contradiction.
Therefore, disregarding the restriction A<= B <= C, we have:
(A, B, C) = (-4,1,1), (1,-4,1), (1,1,-4) as the possible solutions.
Consequently, combining CASES (A) and (B), we have:
(A, B, C) = (1,4,5), (1,5,4), (4,5,1), (4,1,5), (5,1,4), (5,4,1)(1,3,8), (1A,8,3), (3,8,1), (3,1,8), (8,1,3), (8,3,1), (2,2,3), (-4,1,1), (1,-4,1), (1,1,-4) as all possible solutions corresponding to the given problem.
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