Prove the following

Prove that if the polynomial $P(x) = a_0 x^n + a_1 x^{n-1} + \cdots + a_{n-1} x + a_n$ with integral coefficients, takes on the value 7 for four integral values of $x$ , then it cannot have the value 14 for any integral value of $x$ .

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Comments

Let $p(x)=(x-a)(x-b)(x-c)(x-d) + 7$ for $a >b>c>d$

Assume for $x=m$ for which $p(m)=14$

So, $p(m)=14=(m-a)(m-b)(m-c)(m-d) + 7 => (m-a)(m-b)(m-c)(m-d)=7$

If $7$ is factorized (including positive and negative factors) then there will be at least two same factors.

If $x-a=x-c=d$ then it implies $a=c$ which contradicts our assumption.

So, there is no such $x$ such that $p(x)=14$

Kushal Bose - 4 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$