Olympiad Maths Prep

Library / /9 of 28

Algebra Difficulty 5.2 AIME, harder Prove it Ukraine

Does there exist a polynomial f(x)=x3+ax2+bx+cf(x) = x^3 + a x^2 + b x + c satisfying simultaneously the following conditions: ∣c∣≤2009|c| \leq 2009, ff has 3 integer roots and ∣f(34)∣|f(34)| is a prime number.

Solution

Let f(x)=(x−α)(x−β)(x−γ)f(x) = (x - \alpha)(x - \beta)(x - \gamma) be a polynomial which satisfies the necessary conditions. Then α,β,γ\alpha, \beta, \gamma are integer numbers and ∣f(34)∣=∣(34−α)(34−β)(34−γ)∣|f(34)| = |(34 - \alpha)(34 - \beta)(34 - \gamma)| is a prime number. Without loss of generality we have ∣34−α∣=∣34−β∣=1|34 - \alpha| = |34 - \beta| = 1 and ∣34−γ∣|34 - \gamma| is a prime, it follows α,β≥33\alpha, \beta \geq 33. The nearest prime numbers for 3434 are 3131 and 3737 then ∣γ∣≥3|\gamma| \geq 3.
Therefore ∣c∣=∣αβγ∣≥33×33×3=3267>2009|c| = |\alpha \beta \gamma| \geq 33 \times 33 \times 3 = 3267 > 2009. A contradiction.

Looking for a route rather than an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.