For some silly integer , the polynomial has the three integer roots and . Find .
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With Vieta's formula , we know that a + b + c = 0 , and a b + b c + a c = − 2 0 1 1 .
a , b , c = 0 since any one being zero will make the the other 2 ± 2 0 1 1 . a = − ( b + c ) . WLOG, let ∣ a ∣ ≥ ∣ b ∣ ≥ ∣ c ∣ . Then if a > 0 , then b , c < 0 and if a < 0 , b , c > 0 .
a b + b c + a c = − 2 0 1 1 = a ( b + c ) + b c = − a 2 + b c
a 2 = 2 0 1 1 + b c We know that b , c have the same sign. So ∣ a ∣ ≥ 4 5 . ( 4 4 2 < 2 0 1 1 and 4 5 2 = 2 0 2 5 ) Also, b c maximize when b = c if we fixed * *a. Hence, 2 0 1 1 = a 2 − b c > 4 3 a 2 .
So a 2 < 3 ( 4 ) 2 0 1 1 = 2 6 8 1 + 3 1 .
5 2 2 = 2 7 0 4 so ∣ a ∣ ≤ 5 1 .
Now we have limited a to 4 5 ≤ ∣ a ∣ ≤ 5 1 .
Let's us analyze a 2 = 2 0 1 1 + b c .
Here is a table:
∣ a ∣ ⟹ a 2 = 2 0 1 1 + b c
4 5 ⟹ 1 4
4 6 ⟹ 1 4 + 9 1 = 1 0 5
4 7 ⟹ 1 0 5 + 9 3 = 1 9 8
4 8 ⟹ 1 9 8 + 9 5 = 2 9 3
4 9 ⟹ 2 9 3 + 9 7 = 3 9 0
We can tell we don't need to bother with 45, 1 0 5 = ( 3 ) ( 5 ) ( 7 ) , So 46 won't work. 1 9 8 / 4 7 > 4 , 198 is not divisible by 5, 198/6 = 33, which is too small to get 47. 293/48 > 6, 293 is not divisible by 7 or 8 or 9, we can clearly tell that 10 is too much.
Hence, ∣ a ∣ = 4 9 , a 2 − 2 0 1 1 = 3 9 0 . b = 3 9 , c = 1 0 .
Answer: 9 8