Iranian Mathematical Olympiad

Question

Let 1 < t < 2 be a real number. Prove that for all sufficiently large positive integers like d, there is a monic polynomial P(x) of degree d, such that all of its coefficients are either +1 or -1 and |P(t) - 2019| < 1.

Accepted Answer

At first we shall prove following lemma: Lemma. Let

b_{n}

be a sequence of positive real numbers satisfying

b_{n} \leq 2 b_{0} + b_{1} + \dots + b_{n - 1},

then for each real number

z

where

∣ z ∣ \leq 2 b_{0} + b_{1} + \dots + b_{n},

there are

a_{0}, a_{1}, \dots, a_{n} \in {1, - 1}

such that

z - i = 0 \sum n a_{i} b_{i} < b_{0} .

Proof. Write the inequality

∣ z ∣ \leq 2 b_{0} + b_{1} + \dots + b_{n}

in the form

∣ z - sgn (z) b_{n} ∣ \leq 2 b_{0} + b_{1} + \dots + b_{n - 1},

then, proceed the proof by induction on

n

. For sake of convenience, we also define

sgn (0) = 1

. This completes our proof.

□

Back to the problem, define

b_{i} = t^{i}

, then it is easy to deduce that

b_{n} - b_{0} = t^{n} - 1 \leq 1 + t + \dots + t^{n - 1} = \frac{t ^{n} - 1}{t - 1} = b_{0} + b_{1} + \dots + b_{n - 1} .

Moreover, choose

d

such that

t^{d} \geq 2019

, then

2019 \leq t^{d} \leq 2 + t + \dots + t^{d} .

Hence, by the lemma, there are

a_{0}, a_{1}, \dots, a_{d} \in {1, - 1}

such that

i = 0 \sum d a_{i} t^{i} - 2019 < 1.

We are done.