Duplicate roots everywhere
Let x , y , z be all distinct real numbers such that k = x + y + z = x 2 + y 2 + z 2 = x 3 + y 3 + z 3
If the solution set of k is ( a 1 , a 2 ) ∪ ( a 3 , a 4 ) ∪ ⋯ ∪ ( a n − 1 , a n ) , where a 1 < a 2 < … < a n , find a 1 + a 2 + ⋯ + a n .
Bonus:
Can we generalize this for
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The answer is 3.
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1 solution
Using @MarkHenning's method, we can try to generalize. I conjecture that for general n>1 the discriminant will look like
c (k - n)^(n-1) k^(n-1) (k - 1)^(n-2) (k-2)^(n-2) ... (k - (n-1))^(n-2)
where c is a constant that is positive if n is 0 or 3 (mod 4) and negative otherwise. Note that based on the sign of c and the multiplicities of the roots of the discriminant, we can easily determine the sign of the discriminant at any k.
It is also easy to show with the plane and the sphere that -n < k < n. We could probably also show that k > 0.
So using the above discriminant (doing casework for each remainder mod 4) we could easily find a subset S of (0,n) which is a superset of the solution set of k.
The reason it is a superset is because the discriminant of a polynomial is positive iff the number of nonreal zeros is a multiple of 4. So for example if n = 4, S may contain k where all a1, a2, a3, a4 are nonreal.
Note that S will consist of intervals of the form (j,j+1) for some nonnegative integer j < n. Further, note that if any k in (j,j+1) corresponds to a solution (i.e., a1,..., an are all real) then the entire interval (j,j+1) will correspond to real solutions because we can only transition from real solutions to complex solutions by having a multiple root.
By looking at which intervals (j,j+1) correspond to all real roots in small cases, we may be able to find a pattern.
Using Vieta's formulae, we obtain that x y + x z + y z = 2 1 k ( k − 1 ) x y z = 6 1 k ( k − 1 ) ( k − 2 ) and hence x , y , z are the real roots of the cubic polynomial X 3 − k X 2 + 2 1 k ( k − 1 ) X − 6 1 k ( k − 1 ) ( k − 2 ) = 0 This cubic polynomial has cubic discriminant Δ = − 3 k 2 + 2 1 3 k 3 − 6 2 9 k 4 + 2 3 k 5 − 6 1 k 6 = − 6 1 k 2 ( k − 1 ) ( k − 2 ) ( k − 3 ) 2 Thus Δ > 0 when k ∈ ( 1 , 2 ) , while Δ ≥ 0 when k ∈ { 0 } ∪ [ 1 , 2 ] ∪ { 3 } . Distinct real numbers x , y , z (or, equivalently, noninteger k ) can be found with the desired conditions provided that k ∈ ( 1 , 2 ) . Thus the answer is 1 + 2 = 3 .