Question-213641

Question Number 213641 by Abdullahrussell last updated on 12/Nov/24

Answered by Ghisom last updated on 03/Dec/24

α_{k} \in Z

p (x) = \underset{k = 3}{\sum^{n}} α_{k} x^{k}

x \in Z \Rightarrow p (x) \in Z

{we}^{'} ll need this later :

2 ∣ p q \Rightarrow 2 ∣ p \lor 2 ∣ q

2 ∤ p q \Rightarrow 2 ∤ p \land 2 ∤ q

\Rightarrow

2 ∣ α_{k} x^{k} \Rightarrow 2 ∣ α_{k} {(x + 2)}^{k}

2 ∤ α_{k} x^{k} \Rightarrow 2 ∤ α_{k} {(x + 2)}^{k}

\Rightarrow (2 ∣ p (x) \land 2 ∣ p (x + 2)) \lor (2 ∤ p (x) \land 2 ∤ p (x + 2)) ★

f (x) = p (x) + {a x}^{2} + b x + c

inserting f (1) = 2 \land f (2) = 3 \land f (3) = 5 and

solving for a, b, c we get

a = \frac{1}{2} - \frac{p (1)}{2} + p (2) - \frac{p (3)}{2}

b = - \frac{1}{2} + \frac{5 p (1)}{2} - 4 p (2) + \frac{3 p (3)}{2}

c = 2 - 3 p (1) + 3 p (2) - p (3) \Rightarrow c \in Z

a \in Z \Rightarrow \frac{1}{2} - \frac{p (1)}{2} - \frac{p (3)}{2} = M \in Z

b \in Z \Rightarrow - \frac{1}{2} + \frac{5 p (1)}{2} + \frac{3 p (3)}{2} = N \in Z

\Rightarrow

p (1) = 3 M + N - 1

p (3) = - 5 M - N + 2

all combinations of M and N being odd or

even yield

2 ∣ p (1) \Leftrightarrow 2 ∤ p (3) \land 2 ∤ p (1) \Leftrightarrow 2 ∣ p (3)

but

p (3) = p (1 + 2) \Rightarrow contradiction to ★

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