2019 Polynomial problem

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2019 Polynomial problem

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srnjbr

#1 h Mar 25, 2025, 6:15 PM

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suppose t is a member of the interval (1,2). show that there exists a polynomial p with coefficients +-1 such that |p(t)-2019|<=1

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pco

#2 h Mar 27, 2025, 11:19 AM

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srnjbr wrote:

suppose t is a member of the interval (1,2). show that there exists a polynomial p with coefficients +-1 such that |p(t)-2019|<=1

Preliminary claim : $\forall A\in[0,t^n)$ ( $n$ positive integer), $\exists Q(x)$ polynomial with coefficients $0,1$ and degree $<n$ such that $A-Q(t)\in[0,t)$
Proof

Let then $n$ such that $\sum_{i=0}^nt^i\ge 2018>\sum_{i=0}^{n-1}t^i$
Let $A=\frac{\sum_{i=0}^nt^i-2018}2$ : $A\in[0,\frac{t^n}2)\subset [0,t^{n-1})$
And so, using claim, $\exists Q(x)$ polynomial with coefficients $0,1$ and degree $<n-1$ such that $A-Q(t)\in[0,t)$
So $2A-2Q(t)\in[0,2t)$
Which is $\sum_{i=0}^nt^i-2018-2Q(t)\in[0,2t)$
$\implies$ $\sum_{i=0}^nt^i-2Q(t)\in[2018,2018+2t)\subset[2018,2020]$
$\implies$ $\sum_{i=0}^nt^i-2Q(t)-2019\in[-1,1]$
And $P(x)=\sum_{i=0}^nx^i-2Q(x)$ is a polynomial whose all coefficients are $\pm 1$

Q.E.D

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srnjbr

#3 h Mar 30, 2025, 12:58 PM

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THANK YOU

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