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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 1 year, 3 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 \(p(x)=14\)

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`*italics*`

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TopNewestLet \(p(x)=(x-a)(x-b)(x-c)(x-d) + 7\) 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 \(p(x)=14\)

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