Each of P, Q, R, and S is a positive integer with P<Q<R<S.
Find the quadruplets (P, Q, R, S) that satisfy this equation:
1/P + 1/Q + 1/R + 1/S = 1
Prove that these are the only possible quadruplets that satisfy the given conditions.
Note: Computer program solutions are welcome, but a semi-analytical solution is preferred.
Let M = 1/P + 1/Q + 1/R + 1/S.
If P > 2, then the maximum value of M = 1/P + 1/Q + 1/R + 1/S = 1/3 + 1/4 + 1/5+ 1/6 = 57/60 < 1. On the other hand, if P = 1, then M > 1 as well.
Therefore any valid quadruplets must have P = 2.
Given P = 2, 1/Q + 1/R + 1/S = 1/2. Since Q < R < S, Q < 6.
Given P = 2, if Q = 5, then 1/P + 1/Q = 7/10. So we must have 1/R + 1/S = 3/10, with 5 < R < S. Furthermore, since R < S, R < 20/3, otherwise 1/R + 1/S will be less than 3/10. So the only possible value of R is 6. However, if R = 6, then S = 15/2, which is not an integer. So there are no solutions with (P, Q) = (2, 5).
Given P = 2, if Q = 4, then 1/R + 1/S = 1/4. Therefore 4 < R, and since R < S, we also have R < 8. So there are three cases to check:
(P, Q, R) = (2, 4, 5) => S = 20
(P, Q, R) = (2, 4, 6) => S = 12
(P, Q, R) = (2, 4, 7) => S = 28/3
So there are two solutions in this case (P, Q, R, S) = (2, 4, 5, 20) and (2, 4, 6, 12)
Given P = 2, if Q = 3, then 1/R + 1/S = 1/6, so 6 < R. We must also have R < 12, so we have five cases to check here:
(P, Q, R) = (2, 3, 7) => S = 42
(P, Q, R) = (2, 3, 8) => S = 24
(P, Q, R) = (2, 3, 9) => S = 18
(P, Q, R) = (2, 3, 10) => S = 15
(P, Q, R) = (2, 3, 11) => S = 66/5
So in this case we have an additional four solutions: (P, Q, R, S) = (2, 3, 7, 42) , (2, 3, 8, 24), (2, 3, 9, 18), and (2, 3, 10, 15).
Since we have exhausted all the possibilities, these are the only six solutions.
Edited on February 26, 2023, 2:53 pm
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Posted by H M
on 2023-02-24 08:55:09 |
Solution
