α \alpha makes the difference

Suppose n 0 n \geq 0 is an integer and all the roots of x 3 + α x + 4 ( 2 × 201 6 n ) = 0 x^3+ \alpha x + 4 - (2 × 2016^n) = 0 are integers. Find the sum of all possible values of α \alpha .


Source : INMO


The answer is -3.

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1 solution

Patrick Corn
Nov 24, 2017

Let a , b , c a,b,c be the roots of x 3 + α x + 4 ( 2 × 201 6 n ) . x^3+\alpha x + 4 - (2 \times 2016^n). Then a + b + c = 0 a+b+c = 0 and a b c = 2 × 201 6 n 4. abc = 2 \times 2016^n - 4.

Suppose first that n 1. n \ge 1. Then, since 7 2016 , 7|2016, we get a b c 3 abc \equiv 3 mod 7. 7. Since c = a b , c = -a-b, this becomes a b ( a + b ) 4 ab(a+b) \equiv 4 mod 7. 7. Surprisingly enough, this equation has no solutions!

On the other hand, when n = 0 , n=0, a + b + c = 0 a+b+c=0 and a b c = 2. abc = -2. It's not hard to see that the unique solution (up to reordering) over the integers is a = 1 , b = 1 , c = 2 , a=1,b=1,c=-2, which leads to α = a b + b c + a c = 3. \alpha = ab+bc+ac=-3. So the answer is -3 . \fbox{-3}.

(A fancy explanation for why a b ( a + b ) 4 ab(a+b) \equiv 4 mod 7 7 has no solutions is: notice that 3 a b ( a + b ) = ( a + b ) 3 a 3 b 3 , 3ab(a+b) = (a+b)^3-a^3-b^3, and the cubes mod 7 7 are 0 , ± 1. 0,\pm 1. The only ways to get 2 -2 from adding three cubes are if they are 0 , 1 , 1 , 0,-1,-1, or 1 , 0 , 1 , -1,0,-1, or 1 , 1 , 0. -1,-1,0. The former is impossible because a + b = 0 a+b=0 means that a 3 a^3 and b 3 b^3 have opposite signs. The last two are impossible because if e.g. a = 0 a=0 then ( a + b ) 3 b 3 0 (a+b)^3-b^3 \equiv 0 . So 3 a b ( a + b ) 2 3ab(a+b) \equiv -2 is impossible, so a b ( a + b ) 4 ab(a+b) \equiv 4 is impossible.)

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