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# 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 $$x$$.

Note by Himanshu Tiwari
7 months ago

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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 $$p(x)=14$$

- 6 months, 3 weeks ago