Suppose is an integer and all the roots of are integers. Find the sum of all possible values of .
Source : INMO
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Let a , b , c be the roots of x 3 + α x + 4 − ( 2 × 2 0 1 6 n ) . Then a + b + c = 0 and a b c = 2 × 2 0 1 6 n − 4 .
Suppose first that n ≥ 1 . Then, since 7 ∣ 2 0 1 6 , we get a b c ≡ 3 mod 7 . Since c = − a − b , this becomes a b ( a + b ) ≡ 4 mod 7 . Surprisingly enough, this equation has no solutions!
On the other hand, when n = 0 , a + b + c = 0 and a b c = − 2 . It's not hard to see that the unique solution (up to reordering) over the integers is a = 1 , b = 1 , c = − 2 , which leads to α = a b + b c + a c = − 3 . So the answer is - 3 .
(A fancy explanation for why a b ( a + b ) ≡ 4 mod 7 has no solutions is: notice that 3 a b ( a + b ) = ( a + b ) 3 − a 3 − b 3 , and the cubes mod 7 are 0 , ± 1 . The only ways to get − 2 from adding three cubes are if they are 0 , − 1 , − 1 , or − 1 , 0 , − 1 , or − 1 , − 1 , 0 . The former is impossible because a + b = 0 means that a 3 and b 3 have opposite signs. The last two are impossible because if e.g. a = 0 then ( a + b ) 3 − b 3 ≡ 0 . So 3 a b ( a + b ) ≡ − 2 is impossible, so a b ( a + b ) ≡ 4 is impossible.)