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Question-213641




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 ★
αkZp(x)=nk=3αkxkxZp(x)Zwellneedthislater:2pq2p2q2pq2p2q2αkxk2αk(x+2)k2αkxk2αk(x+2)k(2p(x)2p(x+2))(2p(x)2p(x+2))f(x)=p(x)+ax2+bx+cinsertingf(1)=2f(2)=3f(3)=5andsolvingfora,b,cwegeta=12p(1)2+p(2)p(3)2b=12+5p(1)24p(2)+3p(3)2c=23p(1)+3p(2)p(3)cZaZ12p(1)2p(3)2=MZbZ12+5p(1)2+3p(3)2=NZp(1)=3M+N1p(3)=5MN+2allcombinationsofMandNbeingoddorevenyield2p(1)2p(3)2p(1)2p(3)butp(3)=p(1+2)contradictionto

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