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 Product = Sum + 2 (Posted on 2010-09-14)
Determine all possible triplet(s) (p, q, r) of positive real numbers, with p ≤ q ≤ r, that satisfy the following system of equations:

p*q + q*r + p*r = 12 , and:
p*q*r = p + q + r + 2

 No Solution Yet Submitted by K Sengupta No Rating

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 solution | Comment 1 of 3

Starting with p*q*r = p + q + r + 2:

We need not consider any set where p>2, as 3*3*r = 9*r > 8+r = 3+3+r+2 for r>=3.

For any given p and q, r = (p + q + 2) / (p * q - 1), and for any given p this is monotonically decreasing as q increases, once q >= 2:

`p  q  r(calculated)1  1 inf1  1  01  2  51  3  31  4  2.3333331  5  21  6  1.81  7  1.6666671  8  1.571429`
`2  2  22  3  1.42  4  1.1428572  5  12  6  .90909092  7  .84615392  8  .82  9  .7647059`

Beyond this, the r values are fractional, and certainly less than q.

The only candidates are (1,2,5), (1,3,3) and (2,2,2).

1*2 + 2*5 + 1*5 = 17
1*3 + 3*3 + 1*3 = 15
2*2 + 2*2 + 2*2 = 12

So (2,2,2) is the only triplet that satisfies both conditions.

 Posted by Charlie on 2010-09-14 17:28:10

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