Christmas Streak 02/88: ive ractions atural umber
and are all natural numbers.
Sum up all the possible values for
Notation: denotes the maximum value among all the elements in the function.
This problem is a part of <Christmas Streak 2017> series .
The answer is 52.
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1 solution
Let a + 1 = α , b + 1 = β , c + 1 = γ and d + 1 = δ .
Then, without losing generality we can set α ≥ β ≥ γ ≥ δ > 1 .
α 1 + β 1 + γ 1 + δ 1 − α β γ δ 1 = α β γ δ α β γ + β γ δ + γ δ α + δ α β − 1 = n .
α β γ δ n = α β γ + β γ δ + γ δ α + δ α β − 1 < 4 α β γ .
Therefore δ n < 4 , from which we can infer that n = 1 , δ = 2 , 3 .
If δ = 3 ,
3 α β γ = α β γ + 3 ( α β + β γ + γ α ) − 1
2 α β γ = 3 ( α β + β γ + γ α ) − 1 < 9 α β
Then 3 ≤ γ < 2 9 . But then since 2 α β γ is not a multiple of 3, γ = 4 .
8 α β = 3 ( α β + 4 α + 4 β ) − 1 , 5 α β = 1 2 ( α + β ) − 1 < 2 4 α
Then 4 ≤ β < 5 2 4 , but if β = 4 , 5 α β is a multiple of 4 , which clearly violates 5 α β = 1 2 ( α + β ) − 1 .
Therefore this is not possible.
Then let δ = 2 .
2 α β γ = α β γ + 2 ( α β + β γ + γ α ) − 1
α β γ = 2 ( α β + β γ + γ α ) − 1 < 6 α β .
2 ≤ γ < 6 , but since α β γ is not a multiple of 2, γ = 3 , 5 .
If γ = 5 , 5 α β = 2 ( α β + 5 α + 5 β ) − 1 ,
3 α β = 1 0 ( α + β ) − 1 < 2 0 α .
5 ≤ β < 3 2 0 , but if β = 5 , 6 , then 3 α β is a multiple of 5 , 2 respectively, which violates 3 α β = 1 0 ( α + β ) − 1 .
Therefore, γ = 5 .
If γ = 3 , 3 α β = 2 ( α β + 3 α + 3 β ) − 1 .
Then, α β − 6 ( α + β ) = − 1 .
Solving this most basic form of Diophantine equation, we get
( α − 6 ) ( β − 6 ) = 3 5 .
Since α ≥ β ≥ 3 , we get
( α , β ) = ( 4 1 , 7 ) , ( 1 3 , 1 1 ) .
Therefore, max ( a , b , c , d ) = a = α − 1 = 1 2 or 4 0 .
∴ 1 2 + 4 0 = 5 2 .