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Question-213641

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Question Number 213641 by Abdullahrussell last updated on 12/Nov/24
Answered by Ghisom last updated on 03/Dec/24
α_k ∈Z p(x)=Σ_(k=3) ^n α_k x^k x∈Z ⇒ p(x)∈Z we′ll need this later: 2∣pq ⇒ 2∣p∨2∣q 2∤pq ⇒ 2∤p∧2∤q ⇒ 2∣α_k x^k ⇒ 2∣α_k (x+2)^k 2∤α_k x^k ⇒ 2∤α_k (x+2)^k ⇒ (2∣p(x)∧2∣p(x+2))∨(2∤p(x)∧2∤p(x+2)) ★ f(x)=p(x)+ax^2 +bx+c inserting f(1)=2∧f(2)=3∧f(3)=5 and solving for a, b, c we get a=(1/2)−((p(1))/2)+p(2)−((p(3))/2) b=−(1/2)+((5p(1))/2)−4p(2)+((3p(3))/2) c=2−3p(1)+3p(2)−p(3) ⇒ c∈Z a∈Z ⇒ (1/2)−((p(1))/2)−((p(3))/2)=M∈Z b∈Z ⇒ −(1/2)+((5p(1))/2)+((3p(3))/2)=N∈Z ⇒ p(1)=3M+N−1 p(3)=−5M−N+2 all combinations of M and N being odd or even yield 2∣p(1) ⇔ 2∤p(3) ∧ 2∤p(1) ⇔ 2∣p(3) but p(3)=p(1+2) ⇒ contradiction to ★
$$\alpha_{{k}} \in\mathbb{Z} \\ $$$${p}\left({x}\right)=\underset{{k}=\mathrm{3}} {\overset{{n}} {\sum}}\alpha_{{k}} {x}^{{k}} \\ $$$${x}\in\mathbb{Z}\:\Rightarrow\:{p}\left({x}\right)\in\mathbb{Z} \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{we}'\mathrm{ll}\:\mathrm{need}\:\mathrm{this}\:\mathrm{later}: \\ $$$$\:\:\:\:\:\mathrm{2}\mid{pq}\:\Rightarrow\:\mathrm{2}\mid{p}\vee\mathrm{2}\mid{q} \\ $$$$\:\:\:\:\:\mathrm{2}\nmid{pq}\:\Rightarrow\:\mathrm{2}\nmid{p}\wedge\mathrm{2}\nmid{q} \\ $$$$\:\:\:\:\:\Rightarrow \\ $$$$\:\:\:\:\:\mathrm{2}\mid\alpha_{{k}} {x}^{{k}} \:\Rightarrow\:\mathrm{2}\mid\alpha_{{k}} \left({x}+\mathrm{2}\right)^{{k}} \\ $$$$\:\:\:\:\:\mathrm{2}\nmid\alpha_{{k}} {x}^{{k}} \:\Rightarrow\:\mathrm{2}\nmid\alpha_{{k}} \left({x}+\mathrm{2}\right)^{{k}} \\ $$$$\:\:\:\:\:\Rightarrow\:\left(\mathrm{2}\mid{p}\left({x}\right)\wedge\mathrm{2}\mid{p}\left({x}+\mathrm{2}\right)\right)\vee\left(\mathrm{2}\nmid{p}\left({x}\right)\wedge\mathrm{2}\nmid{p}\left({x}+\mathrm{2}\right)\right)\:\bigstar \\ $$$$ \\ $$$${f}\left({x}\right)={p}\left({x}\right)+{ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\mathrm{inserting}\:{f}\left(\mathrm{1}\right)=\mathrm{2}\wedge{f}\left(\mathrm{2}\right)=\mathrm{3}\wedge{f}\left(\mathrm{3}\right)=\mathrm{5}\:\mathrm{and} \\ $$$$\mathrm{solving}\:\mathrm{for}\:{a},\:{b},\:{c}\:\mathrm{we}\:\mathrm{get} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{{p}\left(\mathrm{1}\right)}{\mathrm{2}}+{p}\left(\mathrm{2}\right)−\frac{{p}\left(\mathrm{3}\right)}{\mathrm{2}} \\ $$$${b}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{5}{p}\left(\mathrm{1}\right)}{\mathrm{2}}−\mathrm{4}{p}\left(\mathrm{2}\right)+\frac{\mathrm{3}{p}\left(\mathrm{3}\right)}{\mathrm{2}} \\ $$$${c}=\mathrm{2}−\mathrm{3}{p}\left(\mathrm{1}\right)+\mathrm{3}{p}\left(\mathrm{2}\right)−{p}\left(\mathrm{3}\right)\:\Rightarrow\:{c}\in\mathbb{Z} \\ $$$${a}\in\mathbb{Z}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{{p}\left(\mathrm{1}\right)}{\mathrm{2}}−\frac{{p}\left(\mathrm{3}\right)}{\mathrm{2}}={M}\in\mathbb{Z} \\ $$$${b}\in\mathbb{Z}\:\Rightarrow\:−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{5}{p}\left(\mathrm{1}\right)}{\mathrm{2}}+\frac{\mathrm{3}{p}\left(\mathrm{3}\right)}{\mathrm{2}}={N}\in\mathbb{Z} \\ $$$$\Rightarrow \\ $$$${p}\left(\mathrm{1}\right)=\mathrm{3}{M}+{N}−\mathrm{1} \\ $$$${p}\left(\mathrm{3}\right)=−\mathrm{5}{M}−{N}+\mathrm{2} \\ $$$$\mathrm{all}\:\mathrm{combinations}\:\mathrm{of}\:{M}\:\mathrm{and}\:{N}\:\mathrm{being}\:\mathrm{odd}\:\mathrm{or} \\ $$$$\mathrm{even}\:\mathrm{yield} \\ $$$$\mathrm{2}\mid{p}\left(\mathrm{1}\right)\:\Leftrightarrow\:\mathrm{2}\nmid{p}\left(\mathrm{3}\right)\:\wedge\:\mathrm{2}\nmid{p}\left(\mathrm{1}\right)\:\Leftrightarrow\:\mathrm{2}\mid{p}\left(\mathrm{3}\right) \\ $$$$\mathrm{but} \\ $$$${p}\left(\mathrm{3}\right)={p}\left(\mathrm{1}+\mathrm{2}\right)\:\Rightarrow\:\mathrm{contradiction}\:\mathrm{to}\:\bigstar \\ $$

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