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Suppose $f(x)$ is a polynomial with integer coefficient. If there exist $a$, $b$, $c$, $d$ such that $f(a)=f(b)=f(c)=f(d)=5$ then there does not exist $k$ such that $f(k)=8$. Prove.

We consider $Q(x)= f(x)-5$. Then $Q(x)$ has 4 distinct roots. According to the problem given, $Q(k)=3$. Hence $(k-a)(k-b)(k-c)(k-d)r(k)=3$. Hence we can not find any contradiction!

Manglu
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  • $(k-a)(k-b)(k-c)(k-d)$ is a product of four distinct integers. Can that divide $3$? – Daniel Fischer Nov 02 '15 at 13:43
  • @lulu Nice find, with such descriptive title... –  Nov 02 '15 at 16:24
  • @NormalHuman I know, right? It caught my eye when it was posted before...odd that they match so exactly. I mean, the argument involved is quite general, depending only on the fact that $8-5$ is a prime. The problem in this form must appear somewhere else, newspaper challenge or some such. – lulu Nov 02 '15 at 16:30

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