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Let $f(x)$ be a polynomial with integer coefficients. Suppose that there exist distinct integers $a_1,a_2,a_3,a_4,$ such that $f(a_1)=f(a_2)=f(a_3)=f(a_4)=3.$ Then show that there does not exist any integer $b$ with $f(b)=14.$

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We have $f(x)=3+(x-a_1)(x-a_2)(x-a_3)(x-a_4)g(x)$ for some polynomial $g$ with integer coefficients. But then $(b-a_1)(b-a_2)(b-a_3)(b-a_4)g(b)=11$. Use the primality of $11$ to get a contradiction.

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  • $\begingroup$ Prime of $11$ is $1$ and $11$. So four distinct factor of $11$ is not possible. Am I need to conclude in this way? $\endgroup$ – Argha May 17 '13 at 18:25
  • $\begingroup$ @Argha that is correct. $\endgroup$ – 6005 May 17 '13 at 18:26
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    $\begingroup$ @Argha: You should be careful. Factors of $11$ are $\pm 1$ and $\pm 11$. That is why the question assumes four distinct integers. $\endgroup$ – 23rd May 17 '13 at 18:38
  • $\begingroup$ @Landscape: Thank you for your comment it not only complete the answer but also completely clear my conception.(+1) $\endgroup$ – Argha May 17 '13 at 18:40
  • $\begingroup$ @Argha: You are welcome! $\endgroup$ – 23rd May 17 '13 at 18:42
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Hint: Consider the polynomial $g(x) = f(x) - 3$. Then $g$ has four known linear factors $(x - a_1), (x - a_2), (x - a_3), (x - a_4)$. What can these four factors evaluate to if $f(b) = 14$?

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  • $\begingroup$ I will be thankful to you if you explain a little bit. $\endgroup$ – Argha May 17 '13 at 18:23

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